Separate the variables of the following differential equation. y s i n 2 ⁡ θ d y d θ = 1 ysin^2⁡\theta\frac{dy}{d\theta}=1

y ∙ d y = c s c 2 θ∙dθ ⁡ y\bullet dy=csc^2\operatorname{\theta\bullet d\theta}

y ∙ d y = s e c 2 θ∙dθ ⁡ y\bullet dy=sec^2\operatorname{\theta\bullet d\theta}

d y = c s c 2 θ∙dθ ⁡ dy=csc^2\operatorname{\theta\bullet d\theta}

d y = s e c 2 θ∙dθ ⁡ dy=sec^2\operatorname{\theta\bullet d\theta}

Find the general solution to the differential equation. d y d x = y x \frac{dy}{dx}=y\sqrt{x}

y = C e 2 3 x 3 2 y=Ce^{\frac23x^{\frac32}}

y = 2 3 x 3 2 + C y=\frac23x^{\frac32}+C

y = 3 2 x 3 2 + C y=\frac32x^{\frac32}+C

y = C e 2 3 x 2 3 y=Ce^{\frac23x^{\frac23}}

Find the particular solution that satisfies the given initial condition d y d x = sin ⁡ x ∙ sec ⁡ y , y ( π 2 ) = π 4 \frac{dy}{dx}=\sin x\bullet\sec y,y\left(\frac{\pi}{2}\right)=\frac{\pi}{4} .

y = s i n − 1 ( − cos ⁡ x + 2 2 ) y=sin^{-1}\left(-\cos x+\frac{\sqrt2}{2}\right)

y = s i n − 1 ( − cos ⁡ x + π 4 ) y=sin^{-1}\left(-\cos x+\frac{\pi}{4}\right)

y = − cos ⁡ x + 2 2 y=-\cos x+\frac{\sqrt2}{2}

y = − cos ⁡ x + π 4 y=-\cos x+\frac{\pi}{4}

13. Intro to Differential Equations

Separable Differential Equations

13. Intro to Differential Equations

Separable Differential Equations: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Separable differential equations can be solved by separating variables and integrating both sides. The general form is $d y / d x = f (x) * g (y)$ . Applications include exponential growth and decay, modeled by $d y / d t = k * y$ , and Newton's law of cooling, expressed as $d T / d t = k (T - T_s)$ . Understanding these concepts is crucial for solving real-world problems involving rates of change.

concept

Separation of Variables

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Separation of Variables Video Summary

In the study of differential equations, a first-order differential equation is classified as separable when it can be expressed in the form $ \frac{dy}{dx} = f(x) \cdot g(y) $, where $ f(x) $ is a function of $ x $ and $ g(y) $ is a function of $ y $. This structure allows us to isolate the variables, making it easier to solve the equation. For instance, consider the equation $ \frac{dy}{dx} = (x^2 - 3)(6y^3) $. Here, the right side is clearly a product of a function of $ x $ and a function of $ y $.

To solve separable differential equations, the first crucial step is to separate the variables. This involves rearranging the equation so that all terms involving $ y $ and $ dy $ are on one side, while all terms involving $ x $ and $ dx $ are on the other side. For example, starting with $ \frac{dy}{dx} = (x^2 - 3)(6y^3) $, we can divide both sides by $ 6y^3 $ and multiply both sides by $ dx $ to achieve the separation:

$ \frac{1}{6y^3} dy = (x^2 - 3) dx $

In another example, the equation $ \sqrt{x} - \sqrt{y} \frac{dy}{dx} = 0 $ can be rearranged by isolating $ dy $ and $ dx $. By moving $ \sqrt{x} $ to the other side and manipulating the equation, we can express it as:

$ \sqrt{y} dy = \sqrt{x} dx $

Similarly, for the equation $ 2x \frac{dy}{dx} - y \ln(x^3) = 0 $, we can rewrite it as $ 2x \frac{dy}{dx} = y \ln(x^3) $. By dividing both sides by $ 2x $ and multiplying by $ dx $, we can separate the variables:

$ \frac{dy}{y} = \frac{\ln(x^3)}{2x} dx $

Mastering the skill of separating variables is essential for progressing to the next steps in solving differential equations. This foundational technique relies heavily on algebraic manipulation, and practicing various examples will enhance your proficiency in this area.

example

Separation of Variables Example 1

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Separation of Variables Example 1 Video Summary

In solving differential equations, one common technique is to separate the variables, allowing us to isolate the differentials on opposite sides of the equation. This method is particularly useful when dealing with equations that can be expressed in a form where one side contains only the variable and its differential, while the other side contains the other variable and its differential.

Consider the first example where we have the equation:

\[ \frac{dr}{dt} = e^{r - 2t} \]

Using the properties of exponents, we can rewrite this as:

\[ \frac{dr}{dt} = \frac{e^r}{e^{2t}} \]

This allows us to separate the variables by multiplying both sides by $\frac{1}{e^r}$ and $dt$, leading to:

\[ \frac{1}{e^r} dr = \frac{1}{e^{2t}} dt \]

Here, we have successfully separated the variables, demonstrating that a differential equation is separable if it can be expressed as a product of functions of the two variables.

In the second example, we start with:

\[ y' - 3 + x y^2 - 4 = 0 \]

Rewriting $y'$ as $\frac{dy}{dx}$ gives us:

\[ \frac{dy}{dx} = 3 + x(y^2 - 4) \]

To separate the variables, we can rearrange this to:

\[ \frac{dy}{y^2 - 4} = (3 + x) dx \]

This clearly shows that we can isolate $dy$ and $dx$ on opposite sides, confirming that this equation is separable.

In contrast, the third example:

\[ \frac{dy}{dx} = x + e^y + y^2 \]

does not lend itself to separation. The presence of mixed terms (addition of functions of $x$ and $y$) prevents us from isolating the differentials effectively, indicating that this equation is not separable.

Finally, consider the equation:

\[ t^2 dg - \sqrt{g} - 70 dt = 0 \]

By rearranging, we can express it as:

\[ t^2 dg = \sqrt{g} - 70 dt \]

Dividing both sides appropriately allows us to separate the variables:

\[ \frac{1}{\sqrt{g} - 70} dg = \frac{1}{t^2} dt \]

This shows that even when the equation is presented differently, it can still be manipulated to achieve separability.

In summary, recognizing when a differential equation can be separated is crucial. It often involves rewriting terms and applying algebraic manipulation to isolate the differentials, allowing for easier integration and solution of the equation.

Problem

Separate the variables of the following differential equation.
$y s i n^{2} θ \frac{d y}{d θ} = 1$

$y ∙ d y = s e c^{2} θ∙dθ$

$y ∙ d y = c s c^{2} θ∙dθ$

$d y = c s c^{2} θ∙dθ$

$d y = s e c^{2} θ∙dθ$

concept

Solving Separable Differential Equations

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Solving Separable Differential Equations Video Summary

Separable differential equations can be effectively solved by following a systematic approach that involves separating variables and integrating both sides of the equation. To illustrate this process, consider the differential equation $ y' = y(x + 2) $, which can be rewritten as $ \frac{dy}{dx} = y(x + 2) $. The first step is to separate the variables, which involves isolating the function of $ y $ and $ dy $ on one side and the function of $ x $ and $ dx $ on the other. This can be achieved by dividing both sides by $ y $ and multiplying both sides by $ dx $, resulting in the equation $ \frac{1}{y} dy = (x + 2) dx $.

Next, we integrate both sides. The integral of $ \frac{1}{y} dy $ yields $ \ln |y| $, while the integral of $ (x + 2) dx $ can be computed using the power rule, resulting in $ \frac{1}{2} x^2 + 2x + C $, where $ C $ is the constant of integration. Thus, we have:

$ \ln |y| = \frac{1}{2} x^2 + 2x + C $

To find a particular solution, we utilize an initial condition, such as $ y(-2) = 1 $. Substituting $ x = -2 $ and $ y = 1 $ into the equation allows us to solve for $ C $. Since $ \ln(1) = 0 $, we simplify the equation to find $ C $ by solving:

$ 0 = \frac{1}{2}(-2)^2 + 2(-2) + C $

This simplifies to $ 0 = 2 - 4 + C $, leading to $ C = 2 $. Now, substituting $ C $ back into the equation gives:

$ \ln |y| = \frac{1}{2} x^2 + 2x + 2 $

To isolate $ y $, we exponentiate both sides, resulting in:

$ |y| = e^{\frac{1}{2} x^2 + 2x + 2} $

Since we are looking for a particular solution and have determined $ C $, we can drop the absolute value, leading to:

$ y = e^{\frac{1}{2} x^2 + 2x + 2} $

This process demonstrates how to solve separable differential equations, emphasizing the importance of separating variables, integrating, and applying initial conditions to find particular solutions. In some cases, isolating $ y $ may not be possible, particularly with implicit equations, but this is less common in practice.

example

Solving Separable Differential Equations Example 2

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Solving Separable Differential Equations Example 2 Video Summary

To solve the differential equation $2 + x^2 \frac{dy}{dx} - xy = 0$, we start by rewriting the equation to separate the variables. This involves expressing $y'$ as $\frac{dy}{dx}$, leading to the equation:

$2 + x^2 \frac{dy}{dx} = xy$

Next, we isolate $\frac{dy}{dx}$ by rearranging the equation:

$\frac{dy}{dx} = \frac{xy}{2 + x^2}$

To separate the variables, we multiply both sides by $\frac{1}{y}$ and $dx$, resulting in:

$\frac{1}{y} dy = \frac{x}{2 + x^2} dx$

Now, we can integrate both sides. The left side integrates to:

$\int \frac{1}{y} dy = \ln |y|$

For the right side, we use a substitution where $u = 2 + x^2$, giving $du = 2x dx$ or $\frac{1}{2} du = x dx$. Thus, the integral becomes:

$\int \frac{x}{2 + x^2} dx = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln |u| + C$

Substituting back for $u$, we have:

$\frac{1}{2} \ln |2 + x^2| + C$

Combining the results from both integrals, we get:

$\ln |y| = \frac{1}{2} \ln |2 + x^2| + C$

To solve for $y$, we exponentiate both sides:

\(|y| = e^{C} \sqrt{2 + x^2}\

Letting $k = e^{C}$, we can express the solution as:

$y = k \sqrt{2 + x^2}$

Since $k$ can be any constant (positive or negative), we can simplify this to:

$y = C \sqrt{2 + x^2}$

This represents the general solution to the differential equation.

example

Solving Separable Differential Equations Example 3

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Solving Separable Differential Equations Example 3 Video Summary

To solve the differential equation $ \frac{dy}{dx} = e^{x - y} $ with the initial condition $ y(0) = 2 $, we start by rewriting the equation in a more manageable form. Using the property of exponents, we can express $ e^{x - y} $ as $ \frac{e^x}{e^y} $. This gives us:

$ \frac{dy}{dx} = \frac{e^x}{e^y} $

Next, we separate the variables by multiplying both sides by $ e^y $ and $ dx $, leading to:

$ e^y \, dy = e^x \, dx $

Now, we can integrate both sides. The integral of $ e^y \, dy $ is $ e^y $, and the integral of $ e^x \, dx $ is $ e^x + C $, where $ C $ is the constant of integration. Thus, we have:

$ e^y = e^x + C $

To find the particular solution, we use the initial condition $ y(0) = 2 $. Substituting $ x = 0 $ and $ y = 2 $ into the equation gives:

$ e^2 = e^0 + C $

Since $ e^0 = 1 $, we can simplify this to:

$ e^2 = 1 + C $

Solving for $ C $ yields:

$ C = e^2 - 1 $

Substituting $ C $ back into our equation results in:

$ e^y = e^x + e^2 - 1 $

To isolate $ y $, we take the natural logarithm of both sides:

$ y = \ln(e^x + e^2 - 1) $

This expression represents the particular solution to the differential equation that satisfies the given initial condition. The final solution is:

$ y = \ln(e^x + e^2 - 1) $

Problem

Find the particular solution that satisfies the given initial condition
$(y^{2} + y) e^{x} y^{'} = y^{3} + e^{x} y^{3}; y (0) = 1$

$\ln ∣ y ∣ - \frac{1}{y} = - e^{- x} + x$

$\ln ∣ y ∣ - \frac{1}{y} = - e^{- x} + x + 1$

$\ln ∣ \frac{y}{y^{2}} ∣ = - e^{- x} + x$

$\ln ∣ \frac{y}{y^{2}} ∣ = - e^{- x} + x + 1$

Problem

Find the general solution to the differential equation.
$\frac{d y}{d x} = y \sqrt{x}$

$y = \frac{2}{3} x^{\frac{3}{2}} + C$

$y = \frac{3}{2} x^{\frac{3}{2}} + C$

$y = C e^{\frac{2}{3} x^{\frac{3}{2}}}$

$y = C e^{\frac{2}{3} x^{\frac{2}{3}}}$

Problem

Find the particular solution that satisfies the given initial condition $\frac{d y}{d x} = \sin x ∙ \sec y, y (\frac{π}{2}) = \frac{π}{4}$ .

$y = s i n^{- 1} (- \cos x + \frac{\sqrt{2}}{2})$

$y = s i n^{- 1} (- \cos x + \frac{π}{4})$

$y = - \cos x + \frac{\sqrt{2}}{2}$

$y = - \cos x + \frac{π}{4}$

concept

Exponential Growth & Decay

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Exponential Growth & Decay Video Summary

Exponential growth and decay, also known as exponential change, occurs when the rate of change of a quantity $ y $ with respect to time $ t $ is proportional to the value of $ y $. This relationship can be expressed mathematically as:

\[ \frac{dy}{dt} = k \cdot y \]

Here, $ k $ is a constant that determines the nature of the change. To solve this separable differential equation, we first separate the variables:

\[ \frac{dy}{y} = k \, dt \]

Next, we integrate both sides:

\[ \int \frac{dy}{y} = \int k \, dt \]

This results in:

\[ \ln |y| = kt + C \]

To isolate $ y $, we exponentiate both sides:

\[ |y| = e^{kt + C} = e^{kt} \cdot e^C \]

Letting $ C' = e^C $, we can rewrite this as:

\[ y = C' e^{kt} \]

Since $ C' $ is an arbitrary constant, we can denote it simply as $ C $, leading to the general solution:

\[ y = C e^{kt} \]

In this model, $ y $ represents the quantity experiencing exponential change, $ C $ is the initial value when $ t = 0 $, and $ k $ is the growth constant. The behavior of the system depends on the sign of $ k $: if $ k > 0 $, the system experiences exponential growth, while if $ k < 0 $, it undergoes exponential decay.

For example, consider a population of mice that starts at 100 and grows to 300 in six days. To find the population after three days, we first identify the constants $ C $ and $ k $. Setting $ y(0) = 100 $ gives us:

\[ 100 = C e^{k \cdot 0} \Rightarrow C = 100 \]

Next, using the information that $ y(6) = 300 $, we substitute into the model:

\[ 300 = 100 e^{6k} \]

Dividing both sides by 100 yields:

\[ 3 = e^{6k} \]

Taking the natural logarithm of both sides gives:

\[ \ln(3) = 6k \Rightarrow k = \frac{\ln(3)}{6} \]

Now substituting $ C $ and $ k $ back into the model, we have:

\[ y = 100 e^{\frac{\ln(3)}{6} t} \]

To find the population after three days, we substitute $ t = 3 $:

\[ y(3) = 100 e^{\frac{\ln(3)}{6} \cdot 3} = 100 e^{\frac{3 \ln(3)}{6}} = 100 e^{\frac{\ln(3^3)}{6}} = 100 e^{\ln(3^{1.5})} = 100 \cdot 3^{1.5} \approx 173.2 \]

Rounding to the nearest whole number, we find that there would be approximately 173 mice after three days. This example illustrates how to apply the exponential growth and decay model to real-world scenarios.

example

Exponential Growth & Decay Example 4

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Exponential Growth & Decay Example 4 Video Summary

In this problem, we explore the exponential growth of a population of fruit flies, where we know the population at two different time points: 300 flies on the fourth day and 750 flies on the eighth day. Our goal is to determine the initial population of flies at the start of the experiment.

We define y as the number of fruit flies and t as time in days. The information provided gives us the equations:

y(4) = 300 and y(8) = 750.

Using the exponential growth model, we express the population as:

y(t) = c * e^kt,

where c is the initial population and k is the growth constant. Substituting the known values, we have:

300 = c * e^4k and 750 = c * e^8k.

To eliminate c and solve for k, we divide the two equations:

300 / 750 = (c * e^4k) / (c * e^8k).

This simplifies to:

0.4 = e^{4k - 8k} = e^-4k.

Next, we take the natural logarithm of both sides:

ln(0.4) = -4k.

Solving for k, we find:

k = \frac{ln(0.4)}{-4}.

Now that we have k, we can substitute it back into one of the original equations to find c. Using the first equation:

300 = c * e^4k.

Substituting for k gives us:

300 = c * e^{4 * \frac{ln(0.4)}{-4}}.

This simplifies to:

300 = c * e^-ln(0.4).

Recognizing that e^-ln(0.4) = \frac{1}{0.4}, we can rewrite the equation as:

300 = c * \frac{1}{0.4}.

Multiplying both sides by 0.4 yields:

c = 300 * 0.4 = 120.

Thus, the initial population of fruit flies at the start of the experiment was approximately 120 flies. This problem illustrates how to creatively use algebra and the exponential growth model to find unknown values based on given data.

Problem

The scent of a certain air freshener evaporates at a rate proportional to the amount of the air freshener present. Half of the air freshener evaporates within $2$ hours of being sprayed. If the scent of the air freshener is undetectable once $80 percent sign$ has evaporated, how long will the scent of the air freshener last?

$t equals 1 h of r$

$t equals 2 h of r s$

$t equals 4.64 h of r s$

$t equals 1.83 h of r s$

Problem

$dollar sign 2 comma 000$ is invested in an account that earns interest at a rate of $8.5 percent sign$ and is compounded continuously. Find the particular solution that describes the growth of this account in dollars $A$ after $t$ years. Hint: When interest is compounded continuously, it grows exponentially with a growth constant equivalent to the interest rate.

$A = 2000 e^{0.085 ∙ t}$

$A = 2000 e^{8.5 ∙ t}$

$A = 2000 t e^{0.085}$

$A = 2000 t e^{8.5}$

concept

Newton's Law of Cooling

Video duration:

10m

Newton's Law of Cooling Video Summary

Newton's Law of Cooling describes how the temperature of an object changes over time in relation to the temperature of its surrounding environment. According to this law, the rate of change of an object's temperature, denoted as $ T $, is proportional to the difference between the object's temperature and the ambient temperature, $ T_s $. This relationship can be expressed mathematically as:

\[\frac{dT}{dt} = k(T - T_s)\]

In this equation, $ k $ is a constant that represents the cooling rate, $ T $ is the temperature of the object, and $ T_s $ is the surrounding temperature. To solve problems involving Newton's Law of Cooling, we typically follow a systematic approach.

For example, consider a boiled potato initially at 200°F placed in a room at 68°F. After 10 minutes, its temperature drops to 140°F. We can set up our differential equation by substituting $ T_s = 68 $ into the equation:

\[\frac{dT}{dt} = k(T - 68)\]

Next, we separate the variables and integrate both sides. This leads to:

\[\int \frac{dT}{T - 68} = \int k \, dt\]

Integrating gives us:

\[\ln |T - 68| = kt + C\]

To find the constants $ C $ and $ k $, we use the initial conditions. At $ t = 0 $, $ T(0) = 200 $:

\[\ln |200 - 68| = C \implies C = \ln 132\]

Now, substituting $ C $ back into the equation, we have:

\[\ln |T - 68| = kt + \ln 132\]

Next, we use the second condition, $ T(10) = 140 $, to find $ k $:

\[\ln |140 - 68| = 10k + \ln 132\]

Solving this gives:

\[\ln 72 - \ln 132 = 10k \implies k = \frac{1}{10} \ln \left(\frac{72}{132}\right) = \frac{1}{10} \ln \left(\frac{6}{11}\right)\]

Substituting $ k $ back into the equation, we can express the temperature as a function of time:

\[\ln |T - 68| = \frac{1}{10} \ln \left(\frac{6}{11}\right) t + \ln 132\]

Exponentiating both sides results in:

\[T - 68 = 132 \left(\frac{6}{11}\right)^{\frac{t}{10}}\]

Finally, to find the temperature after 30 minutes, we substitute $ t = 30 $:

\[T(30) = 132 \left(\frac{6}{11}\right)^{3} + 68\]

Calculating this yields approximately 89.42°F. This process illustrates how to apply Newton's Law of Cooling to determine the temperature of an object over time, emphasizing the importance of setting up the differential equation, integrating, and solving for constants based on initial conditions.

example

Newton's Law of Cooling Example 5

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Newton's Law of Cooling Example 5 Video Summary

In this problem, we analyze the cooling of a glass face using Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. The equation can be expressed as:

\[ \frac{dT}{dt} = k(T - T_s) \]

where $ T $ is the temperature of the object, $ T_s $ is the surrounding temperature, and $ k $ is a constant. In this scenario, the glass is initially heated to 1,250 degrees Fahrenheit and is placed in a room at a constant temperature of 74 degrees Fahrenheit. After 10 minutes, the temperature of the glass drops to 900 degrees Fahrenheit.

To solve for the time it takes for the glass to cool to 200 degrees Fahrenheit, we first set up our differential equation with the surrounding temperature:

\[ \frac{dT}{dt} = k(T - 74) \]

Next, we separate the variables and integrate both sides:

\[ \int \frac{1}{T - 74} dT = \int k \, dt \]

This results in:

\[ \ln |T - 74| = kt + C \]

To find the constant $ C $, we use the initial condition $ T(0) = 1250 $:

\[ \ln(1250 - 74) = C \Rightarrow C = \ln(1176) \]

Now, substituting $ C $ back into the equation gives:

\[ \ln |T - 74| = kt + \ln(1176) \]

Next, we need to find the constant $ k $ using the condition $ T(10) = 900 $:

\[ \ln(900 - 74) = 10k + \ln(1176) \]

Solving for $ k $ involves isolating it:

\[ \ln(826) - \ln(1176) = 10k \Rightarrow k = \frac{1}{10} \ln\left(\frac{826}{1176}\right) \]

Now, we substitute $ k $ back into our equation:

\[ \ln |T - 74| = \frac{1}{10} \ln\left(\frac{826}{1176}\right)t + \ln(1176) \]

To find the time when the temperature $ T $ reaches 200 degrees Fahrenheit, we set $ T = 200 $:

\[ \ln(200 - 74) = \frac{1}{10} \ln\left(\frac{826}{1176}\right)t + \ln(1176) \]

Solving this gives:

\[ \ln(126) - \ln(1176) = \frac{1}{10} \ln\left(\frac{826}{1176}\right)t \]

Combining the logarithms results in:

\[ \ln\left(\frac{126}{1176}\right) = \frac{1}{10} \ln\left(\frac{826}{1176}\right)t \]

Multiplying both sides by 10 and isolating $ t $ yields:

\[ t = 10 \cdot \frac{\ln\left(\frac{126}{1176}\right)}{\ln\left(\frac{826}{1176}\right)} \]

Calculating this expression results in approximately 63.22 minutes. Therefore, it takes a little over an hour for the glass to cool down to 200 degrees Fahrenheit.

Problem

A pie is removed from an oven and its temperature is $175 ℃$ and placed into a refrigerator whose temperature is constantly $3 ℃$ . After $1$ hour in the refrigerator, the pie is $90 ℃$ . What is the temperature of the pie $4$ hours after being placed in the refrigerator?

$226.1 ℃$

$214.7 ℃$

$14.26 ℃$

$13.69 ℃$

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What is a separable differential equation and how do you identify it?

A separable differential equation is a first-order differential equation that can be expressed as the product of a function of x and a function of y. Its general form is:

\frac{d y}{d x} = f (x) g (y)

To identify a separable differential equation, check if the equation can be rearranged such that all terms involving y (including dy) are on one side and all terms involving x (including dx) are on the other side. For example, $\frac{d y}{d x} = x^{2} * y^{3}$ is separable because it can be rewritten as:

\frac{d y}{y^{3}} = x^{2} d x

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How do you solve a separable differential equation step by step?

To solve a separable differential equation, follow these steps:

Separate the variables: Rearrange the equation so that all terms involving y and dy are on one side, and all terms involving x and dx are on the other side. For example, $\frac{d y}{d x} = x^{2} y^{3}$ becomes $\frac{d y}{y^{3}} = x^{2} d x$ .
Integrate both sides: Integrate the left side with respect to y and the right side with respect to x. For example: $\int_{y}^{-3} d y = \int_{x^{2}} d x$ .
Simplify the integrals: Solve the integrals to find the general solution. Add a constant of integration C to one side.
Apply initial conditions (if given): Substitute the initial values to find the particular solution.
Solve for y: If possible, isolate y to express the solution explicitly.

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What is the exponential growth and decay model in separable differential equations?

The exponential growth and decay model is a common application of separable differential equations. It describes processes where the rate of change of a quantity is proportional to the quantity itself. The general form is:

\frac{d y}{d t} = k y

Here, k is the growth constant. If k > 0, the process represents exponential growth; if k < 0, it represents exponential decay. Solving this equation involves separating variables:

\frac{d y}{y} = k d t

Integrating both sides gives:

\ln | y | = k t + C

Exponentiating both sides yields:

y = C e^{k t}

where C is the initial value.

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How is Newton's law of cooling modeled using separable differential equations?

Newton's law of cooling describes how the temperature of an object changes over time relative to its surroundings. The differential equation is:

\frac{d T}{d t} = k (T - T_{s)}

Here, T is the object's temperature, Ts is the surrounding temperature, and k is a constant. To solve:

Separate variables: Rearrange to isolate T and dT on one side and t and dt on the other:

\frac{d T}{T - T_{s}} = k d t

Integrate: Solve both sides to find:

\ln | T - T_{s | = k t + C}

Solve for T: Exponentiate and isolate T to express the temperature as a function of time.

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What are the real-world applications of separable differential equations?

Separable differential equations have numerous real-world applications, including:

Exponential Growth and Decay: Used in population modeling, radioactive decay, and compound interest calculations. The model is $\frac{d y}{d t} = k y$ .
Newton's Law of Cooling: Predicts temperature changes of objects in different environments, such as cooling coffee or heating food. The equation is $\frac{d T}{d t} = k (T - T_{s)}$ .
Fluid Dynamics: Models flow rates in pipes and channels.
Chemical Reactions: Describes reaction rates in terms of reactant concentrations.

These equations are essential for understanding dynamic systems in physics, biology, and engineering.

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Separable Differential Equations: Videos & Practice Problems

Separation of Variables

Separation of Variables Video Summary

Separation of Variables Example 1

Separation of Variables Example 1 Video Summary

Separate the variables of the following differential equation.ysin2θdydθ=1ysin^2\theta\frac{dy}{d\theta}=1

Solving Separable Differential Equations

Solving Separable Differential Equations Video Summary

Solving Separable Differential Equations Example 2

Solving Separable Differential Equations Example 2 Video Summary

Solving Separable Differential Equations Example 3

Solving Separable Differential Equations Example 3 Video Summary

Find the particular solution that satisfies the given initial condition (y2+y)exy′=y3+exy3;y(0)=1\left(y^2+y\right)e^{x}y^{\prime}=y^3+e^{x}y^3;y\left(0\right)=1

Find the general solution to the differential equation.dydx=yx\frac{dy}{dx}=y\sqrt{x}

Find the particular solution that satisfies the given initial condition dydx=sinx∙secy,y(π2)=π4\frac{dy}{dx}=\sin x\bullet\sec y,y\left(\frac{\pi}{2}\right)=\frac{\pi}{4}.

Exponential Growth & Decay

Exponential Growth & Decay Video Summary

Exponential Growth & Decay Example 4

Exponential Growth & Decay Example 4 Video Summary

Newton's Law of Cooling

Newton's Law of Cooling Video Summary

Newton's Law of Cooling Example 5

Newton's Law of Cooling Example 5 Video Summary

A pie is removed from an oven and its temperature is 175℃175℃ and placed into a refrigerator whose temperature is constantly 3℃3℃. After 11 hour in the refrigerator, the pie is 90℃90℃. What is the temperature of the pie 44 hours after being placed in the refrigerator?

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What is a separable differential equation and how do you identify it?

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Your Calculus tutors

Separate the variables of the following differential equation.
$y s i n^{2} θ \frac{d y}{d θ} = 1$

Find the particular solution that satisfies the given initial condition
$(y^{2} + y) e^{x} y^{'} = y^{3} + e^{x} y^{3}; y (0) = 1$

Find the general solution to the differential equation.
$\frac{d y}{d x} = y \sqrt{x}$

Find the particular solution that satisfies the given initial condition $\frac{d y}{d x} = \sin x ∙ \sec y, y (\frac{π}{2}) = \frac{π}{4}$ .

A pie is removed from an oven and its temperature is $175 ℃$ and placed into a refrigerator whose temperature is constantly $3 ℃$ . After $1$ hour in the refrigerator, the pie is $90 ℃$ . What is the temperature of the pie $4$ hours after being placed in the refrigerator?