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 \left(y^2+y\right)e^{x}y^{\prime}=y^3+e^{x}y^3;y\left(0\right)=1 ( y 2 + y ) e x y ′ = y 3 + e x y 3 ; y ( 0 ) = 1

ln ⁡ ∣ y ∣ − 1 y = − e − x + x \ln\left|y\right|-\frac{1}{y}=-e^{-x}+x ln ∣ y ∣ − y 1 = − e − x + x

ln ⁡ ∣ y ∣ − 1 y = − e − x + x + 1 \ln\left|y\right|-\frac{1}{y}=-e^{-x}+x+1 ln ∣ y ∣ − y 1 = − e − x + x + 1

ln ⁡ ∣ y y 2 ∣ = − e − x + x \ln\left|\frac{y}{y^2}\right|=-e^{-x}+x ln y 2 y = − e − x + x

ln ⁡ ∣ y y 2 ∣ = − e − x + x + 1 \ln\left|\frac{y}{y^2}\right|=-e^{-x}+x+1 ln y 2 y = − e − x + x + 1

Find the general solution to the differential equation. d y d x = y x \frac{dy}{dx}=y\sqrt{x} d x d y = y x <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

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

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

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

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

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 2 2 hours of being sprayed. If the scent of the air freshener is undetectable once 80 % 80\% 80% has evaporated, how long will the scent of the air freshener last?

t = 4.64 h r s t=4.64hrs t = 4.64 h rs

t = 1.83 h r s t=1.83hrs t = 1.83 h rs

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 $d y / d t = k * y$ , where k indicates growth (positive) or decay (negative). Newton's law of cooling describes temperature change as $d T / d t = k (T - T_s)$ .

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 separate the variables by dividing both sides by $ 6y^3 $ and multiplying by $ dx $, leading to:

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

In another example, consider the equation $ \sqrt{x} - \sqrt{y} \frac{dy}{dx} = 0 $. To separate the variables, we can rearrange it to isolate $ dy $ and $ dx $. By moving $ \sqrt{x} $ to the other side and manipulating the equation, we arrive at:

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

Lastly, in the equation $ 2x \frac{dy}{dx} - y \ln(x^3) = 0 $, we can rewrite $ \frac{dy}{dx} $ as $ y' $ and rearrange the terms to isolate $ dy $ and $ dx $. After some algebraic manipulation, we can express it as:

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

Successfully separating the variables is a fundamental skill in solving separable differential equations, as it sets the stage for further integration and finding solutions. Mastery of this technique is essential before progressing to more complex methods of solving differential equations.

example

Separation of Variables Example 1

Video duration:

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 process can sometimes be less straightforward than it appears at first glance, particularly when dealing with exponential functions or complex expressions.

Consider the equation dr/dt = e^r - 2t. Initially, it may not seem separable, but by applying the rules of exponents, we can rewrite it as dr/dt = e^r / e^2t. This allows us to express the equation in a form that can be separated. By multiplying both sides by 1/e^r and dt, we can rearrange the equation to 1/e^r dr = 1/e^2t dt, successfully separating the variables.

Next, we examine the equation y' - 3 + x*y² - 4 = 0. By rewriting y' as dy/dx and rearranging, we obtain dy/dx = 3 + x(y² - 4). This clearly shows a function of x multiplied by a function of y, indicating that it is separable. Dividing both sides by y² - 4 and multiplying by dx leads to 1/(y² - 4) dy = (3 + x) dx, achieving separation.

In contrast, the equation dy/dx = x + e^y + y² presents a challenge. The presence of addition rather than multiplication between functions of x and y makes it impossible to separate the variables. Attempts to isolate terms do not yield a separable form, confirming that this equation is not separable.

Lastly, consider the equation t² dg - √g - 70 dt = 0. Although it is structured differently, it can still be manipulated into a separable form. By rearranging to t² dg = (√g - 70) dt and dividing both sides appropriately, we arrive at 1/(√g - 70) dg = 1/t² dt, successfully separating the variables.

In summary, recognizing the potential for separation in differential equations often requires creative manipulation of the terms involved. By applying algebraic techniques and understanding the structure of the equations, we can effectively isolate the variables and facilitate the process of solving these equations.

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) $, with the initial condition $ y(-2) = 1 $.

First, rewrite the derivative $ y' $ as $ \frac{dy}{dx} $, leading to the equation:

$ \frac{dy}{dx} = y(x + 2) $

Next, separate the variables by isolating $ y $ and $ dy $ on one side and $ x $ and $ dx $ on the other. This can be achieved by dividing both sides by $ y $ and multiplying both sides by $ dx $:

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

With the variables separated, the next step is to integrate both sides:

$ \int \frac{1}{y} dy = \int (x + 2) dx $

The left side integrates to the natural logarithm of the absolute value of $ y $:

$ \ln |y| $

For the right side, applying the power rule yields:

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

Combining these results gives:

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

To find the particular solution, substitute the initial condition $ y(-2) = 1 $ into the equation. This means plugging in $ x = -2 $ and $ y = 1 $:

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

Since $ \ln(1) = 0 $, the equation simplifies to:

$ 0 = 2 - 4 + C $

Thus, solving for $ C $ gives:

$ C = 2 $

Now substitute $ C $ back into the integrated equation:

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

To isolate $ y $, exponentiate both sides:

$ |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 in a more manageable form. We express $y'$ as $\frac{dy}{dx}$, leading to the equation $2 + x^2 \frac{dy}{dx} = xy$. Next, we isolate $\frac{dy}{dx}$ by rearranging the terms:

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

Now, we separate the variables by moving all terms involving $y$ to one side and all terms involving $x$ to the other side:

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

With the variables separated, 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 $u = 2 + x^2$, which gives $du = 2x dx$ or $\frac{1}{2} du = x dx$. Thus, the integral becomes:

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

Combining the results from both integrals, we have:

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

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

\(|y| = e^{C} \cdot \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, where $C$ is an arbitrary constant determined by initial conditions if provided.

example

Solving Separable Differential Equations Example 3

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Solving Separable Differential Equations Example 3 Video Summary

To solve the separable differential equation given by $ \frac{dy}{dx} = e^{x - y} $ with the initial condition $ y(0) = 2 $, we start by rewriting the equation. Using the property of exponents, we can express $ e^{x - y} $ as $ \frac{e^x}{e^y} $. This allows us to rewrite the equation as:

$ \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 apply 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 initial condition. The final solution is:

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

Problem

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

$\ln\left|y\right|-\frac{1}{y}=-e^{-x}+x$

$\ln\left|y\right|-\frac{1}{y}=-e^{-x}+x+1$

$\ln\left|\frac{y}{y^2}\right|=-e^{-x}+x$

$\ln\left|\frac{y}{y^2}\right|=-e^{-x}+x+1$

Problem

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

$y=\frac23x^{\frac32}+C$

$y=\frac32x^{\frac32}+C$

$y=Ce^{\frac23x^{\frac32}}$

$y=Ce^{\frac23x^{\frac23}}$

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:

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

Since $ e^C $ is a constant, we can denote it as $ c $, leading to the general solution:

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

In this model, $ c $ represents the initial value of $ y $ when $ t = 0 $, and $ k $ is the growth constant. The behavior of the function depends on the sign of $ k $: if $ k > 0 $, the function exhibits exponential growth, while if $ k < 0 $, it represents exponential decay. This can be expressed as two separate equations:

\[y = c e^{kt} \quad \text{(growth)} \quad \text{and} \quad y = c e^{-kt} \quad \text{(decay)}\]

To illustrate this model, consider a scenario where a population of mice increases exponentially. If the initial population is 100 mice and after six days it grows to 300, we can determine the population after three days using the model.

First, we identify the constants $ c $ and $ k $. Since the initial population is 100, we have:

\[y(0) = 100 \implies 100 = c \cdot e^{k \cdot 0} \implies c = 100\]

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

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

Dividing both sides by 100 gives:

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

Taking the natural logarithm of both sides results in:

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

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

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

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

\[y(3) = 100 \cdot e^{\frac{\ln(3)}{6} \cdot 3} = 100 \cdot e^{\frac{3 \ln(3)}{6}} = 100 \cdot e^{\frac{\ln(3^{3/2})}} = 100 \cdot 3^{1/2} = 100 \cdot \sqrt{3} \approx 173.2\]

Rounding to the nearest whole number, we find that there would be approximately 173 mice after three days. This example demonstrates how to apply the exponential growth and decay model to real-world scenarios, allowing for predictions based on initial conditions and growth rates.

example

Exponential Growth & Decay Example 4

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Exponential Growth & Decay Example 4 Video Summary

In this scenario, we are analyzing the exponential growth of a population of fruit flies over time. We know that on the fourth day, the population was 300 flies, and by the eighth day, it had increased to 750 flies. Our goal is to determine the initial population of flies at the start of the experiment.

To model this situation, we define y as the number of fruit flies and t as the time in days. The information provided gives us two equations based on the exponential growth model:

1. $ y(4) = c \cdot e^{4k} = 300 $

2. $ y(8) = c \cdot e^{8k} = 750 $

Here, c represents the initial population, and k is the growth constant. To find c, we first need to solve for k. We can do this by dividing the two equations, which allows us to eliminate c:

$ \frac{300}{750} = \frac{c \cdot e^{4k}}{c \cdot e^{8k}} $

This simplifies to:

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

Next, we take the natural logarithm of both sides to isolate k:

$ \ln(0.4) = -4k $

Solving for k gives us:

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

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

$ 300 = c \cdot e^{4k} $

Substituting for k:

$ 300 = c \cdot e^{4 \cdot \frac{\ln(0.4)}{-4}} $

This simplifies to:

$ 300 = c \cdot e^{-\ln(0.4)} $

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

$ 300 = c \cdot \frac{1}{0.4} $

Multiplying both sides by 0.4 gives:

$ c = 300 \cdot 0.4 = 120 $

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

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 $80\%$ has evaporated, how long will the scent of the air freshener last?

$t=1hr$

$t=2hrs$

$t=4.64hrs$

$t=1.83hrs$

Problem

$\$2,000$ is invested in an account that earns interest at a rate of $8.5\%$ 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.

$A=2000e^{0.085\bullet t}$

$A=2000e^{8.5\bullet t}$

$A=2000te^{0.085}$

$A=2000te^{8.5}$

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 the 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 $, which allows us to solve for $ C $:

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

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

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

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

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

Solving this equation leads to:

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

Substituting $ k $ back into the equation gives us:

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

Exponentiating both sides results in:

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

Finally, solving for $ T $ yields:

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

To find the temperature of the potato after 30 minutes, we substitute $ t = 30 $ into the equation:

\[T(30) = 132 \left(\frac{6}{11}\right)^{3} + 68 \approx 89.42 \text{°F}\]

In summary, when applying Newton's Law of Cooling, start by establishing the differential equation, separate the variables, integrate, and use initial conditions to solve for the constants. This method allows for the determination of an object's temperature at any given time.

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 object 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:

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

Here, $T$ is the temperature of the object, $T_s$ is the surrounding temperature, and $k$ is a constant that represents the cooling rate. 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 plugged in:

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

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 where $T(0) = 1250$:

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

Now, we substitute $C$ back into the equation:

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

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

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

Solving for $k$, we have:

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

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 reaches 200 degrees Fahrenheit, we set $T = 200$:

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

After simplifying, we isolate $t$:

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

Calculating this expression yields approximately 63.22 minutes. Thus, it takes a little over an hour for the glass to cool down to 200 degrees Fahrenheit. This problem illustrates the application of differential equations in real-world scenarios, particularly in understanding heat transfer and cooling processes.

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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We have more practice problems on Separable Differential Equations

Here’s what students ask on this topic:

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 in the form $\frac{d}{y} d x$ = g(x)h(y), where the derivative can be written as the product of a function of x and a function of y. To identify it, check if the equation can be rearranged so that all terms involving y (including dy) are on one side and all terms involving x (including dx) are on the other. For example, $\frac{d}{y} d x$ = x²y³ is separable because it can be rewritten as $\frac{1}{y^{3}} d y$ = x²dx.

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

To solve a separable differential equation, follow these steps:
1. **Separate 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. For example, $\frac{d}{y} d x$ = x²y³ becomes $\frac{1}{y^{3}} d y$ = x²dx.
2. **Integrate Both Sides**: Integrate the y-side and the x-side separately. For example, $\frac{1}{y^{3}} d y$ integrates to $- \frac{1}{2} y^{2}$ , and x²dx integrates to $\frac{1}{3} x^{3}$ + C.
3. **Solve for y**: If possible, isolate y to find the explicit solution. If not, leave the solution in implicit form.

What is the exponential growth and decay model, and how is it derived?

The exponential growth and decay model is given by $y = c e^{k t}$ , where c is the initial value, k is the growth constant, and t is time. It is derived from the separable differential equation $\frac{d}{y} d t$ = ky, which states that the rate of change of y is proportional to y. Separating variables gives $\frac{d}{y} y$ = kdt. Integrating both sides yields $\ln y$ = kt + C. Exponentiating both sides gives $y = e k^{t} e C$ , where $e C$ is replaced by c.

What is Newton's law of cooling, and how do you solve related problems?

Newton's law of cooling states that the rate of change of an object's temperature T is proportional to the difference between its temperature and the surrounding medium's temperature T_s. The equation is $\frac{d}{T} d t$ = k(T - T_s). To solve problems:
1. **Set up the equation**: Substitute the given T_s into the equation.
2. **Separate variables**: Rearrange to $\frac{d}{T} T - T$ _s = kdt.
3. **Integrate**: Solve both sides to find T as a function of t.
4. **Use initial conditions**: Plug in given values to find constants.
5. **Solve for T**: Write the final temperature function and use it to find T at specific times.

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

Separable differential equations have numerous real-world applications, including:
1. **Exponential Growth and Decay**: Used in population modeling, radioactive decay, and compound interest calculations. For example, the population of bacteria growing exponentially can be modeled using $y = c e^{k t}$ .
2. **Newton's Law of Cooling**: Describes how an object cools or heats relative to its environment, such as cooling coffee or heating food.
3. **Mixing Problems**: Models the concentration of substances in a tank with inflow and outflow rates.
4. **Chemical Reactions**: Describes reaction rates proportional to reactant concentrations.
These applications rely on separating variables and integrating to find solutions.

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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

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(y2+y)exy′=y3+exy3;y(0)=1

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

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℃175℃ and placed into a refrigerator whose temperature is constantly 3℃3℃3℃. After 111 hour in the refrigerator, the pie is 90℃90℃90℃. What is the temperature of the pie 444 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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Find the particular solution that satisfies the given initial condition
$\left(y^2+y\right)e^{x}y^{\prime}=y^3+e^{x}y^3;y\left(0\right)=1$

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

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?