State the order of the differential equation and indicate if it is linear or nonlinear. d 2 y d x 2 − ( d y d x ) ( 1 − x ) = 2 \frac{d^2y}{dx^2}-\left(\frac{dy}{dx}\right)(1-x)=2

2 n d 2^{nd} 2 n d order, nonlinear

3 r d 3^{rd} 3 r d order, linear

Find the particular solution to the differential equation y ′ = 2 e t + 4 t y^{\prime}=2e^{t}+4t given the initial condition y ( 0 ) = 1 y\left(0\right)=1 .

y = 2 e t + 2 t 2 − 1 y=2e^{t}+2t^2-1

y = 2 e t + 2 t 2 y=2e^{t}+2t^2

Find the general solution to the differential equation d y d x = − 2 x + 5 x 2 \frac{dy}{dx}=-2x+5x^2 .

y = − x 2 + 5 3 x 3 + C y=-x^2+\frac53x^3+C

y = − x 2 + 10 x + C y=-x^2+10x+C

y = − x + 5 3 x 2 + C y=-x^{}+\frac53x^2+C

Find the particular solution to the differential equation y ′ = 2 sin ⁡ x + 3 cos ⁡ x y^{\prime}=2\sin x+3\cos x given the initial condition y ( 0 ) = 4 y\left(0\right)=4 .

y = − 2 cos ⁡ x + 3 sin ⁡ x + 6 y=-2\cos x+3\sin x+6

y = − 2 cos ⁡ x + 3 sin ⁡ x + 4 y=-2\cos x+3\sin x+4

y = − 2 cos ⁡ x + 3 sin ⁡ x + 1 y=-2\cos x+3\sin x+1

y = − 2 cos ⁡ x + 3 sin ⁡ x y=-2\cos x+3\sin x

13. Intro to Differential Equations

Basics of Differential Equations

13. Intro to Differential Equations

Basics of Differential Equations: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Understanding differential equations (DE) involves recognizing their order, which is determined by the highest derivative present, and classifying them as linear or nonlinear based on specific criteria. A solution to a DE is a function that satisfies the equation when substituted with its derivatives. To find particular solutions, one integrates the DE and applies initial conditions. This process highlights the importance of integration and algebra in solving DEs, enabling students to verify solutions and understand the underlying principles of differential equations.

concept

Classifying Differential Equations

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Classifying Differential Equations Video Summary

In the study of differential equations (DE), we explore equations that involve a function $ y = f(x) $ and its derivatives. Understanding the classification of these equations is crucial, as it helps in solving them effectively. Differential equations can be categorized primarily by their order and linearity. The order of a differential equation is determined by the highest derivative present. For instance, if an equation includes both a first derivative and a second derivative, it is classified as a second-order differential equation.

Another important classification is whether the differential equation is linear or nonlinear. A linear differential equation must satisfy three criteria: first, the dependent variable $ y $ and its derivatives should not be multiplied together; second, $ y $ and its derivatives must only be raised to the first power; and third, $ y $ and its derivatives should not be the argument of any function. If any of these conditions are violated, the equation is considered nonlinear.

For example, consider the differential equation $ y'' + 4y' - 3y = \sin(e^x) $. Here, the highest derivative is the second derivative $ y'' $, making it a second-order differential equation. Since $ y $ and its derivatives are not multiplied together, are raised only to the first power, and are not inside any function, this equation is classified as a second-order linear differential equation.

In contrast, the equation $ t y'' - t^2 y^{(4)} = e^y $ has a highest derivative of the fourth order, thus it is a fourth-order differential equation. However, since $ y $ appears as the argument of the exponential function, it fails the linearity test, making it a fourth-order nonlinear differential equation.

Lastly, consider the equation $ y \frac{dy}{dx} + 5 \left(\frac{dy}{dx}\right)^3 - 4 = 2x $. The highest derivative here is the first derivative $ \frac{dy}{dx} $, indicating it is a first-order differential equation. However, since $ y $ is multiplied by its derivative and the derivative is raised to the third power, this equation is classified as a first-order nonlinear differential equation.

As we continue to explore differential equations, understanding these classifications will enhance our ability to analyze and solve them effectively. Each type of differential equation presents unique challenges and requires specific methods for finding solutions.

Problem

State the order of the differential equation and indicate if it is linear or nonlinear.
$y^{^{'''}} + 3 x y = 4 \sqrt{x}$

$3^{rd}$ order; nonlinear

$1 raised to the s t power$ order; linear

$3 raised to the r d power$ order; linear

$1^{st}$ order; nonlinear

Problem

State the order of the differential equation and indicate if it is linear or nonlinear.
$\frac{d^{2} y}{d x^{2}} - (\frac{d y}{d x}) (1 - x) = 2$

$2 raised to the n d power$ order, linear

$2^{nd}$ order, nonlinear

$3 raised to the r d power$ order, nonlinear

$3^{rd}$ order, linear

Problem

State the order of the differential equation and indicate if it is linear or nonlinear.
${(y^{''})}^{2} + 6 e^{t} y^{'} = 4 t$

$2 raised to the n d power$ order, linear

$2^{nd}$ order, nonlinear

$1 raised to the s t power$ order, linear

$1^{st}$ order, nonlinear

Problem

State the order of the differential equation and indicate if it is linear or nonlinear.
$y^{'} ∙ y = 3 e^{t}$

$1 raised to the s t power$ order, linear

$1^{st}$ order, nonlinear

$2 raised to the n d power$ order, linear

$2^{nd}$ order, nonlinear

concept

Verifying Solutions of Differential Equations

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Verifying Solutions of Differential Equations Video Summary

When working with differential equations, verifying solutions is a crucial skill. A function $ y = f(x) $ is considered a solution to a differential equation if substituting this function and its derivatives into the equation results in a true statement. It's important to note that a differential equation can have multiple solutions, but verification typically involves checking a specific function provided in the problem.

To illustrate this process, consider the function $ y = e^{2x} $ and the differential equation $ 3y' - 5y = e^{2x} $. First, we need to find the first derivative $ y' $. Using the chain rule, we find that $ y' = 2e^{2x} $. Next, we substitute both $ y $ and $ y' $ into the differential equation:

Substituting gives us:

$ 3(2e^{2x}) - 5(e^{2x}) = e^{2x} $

which simplifies to:

$ 6e^{2x} - 5e^{2x} = e^{2x} $

Thus, $ e^{2x} = e^{2x} $, confirming that the function is indeed a solution.

In another example, consider the function $ y = 4 + \ln(x) $ and the differential equation $ x y'' + y' = 0 $. We first find the first and second derivatives. The first derivative is $ y' = \frac{1}{x} $, and the second derivative, using the power rule, is $ y'' = -\frac{1}{x^2} $.

Substituting these into the differential equation yields:

$ x\left(-\frac{1}{x^2}\right) + \frac{1}{x} = 0 $

which simplifies to:

$ -\frac{1}{x} + \frac{1}{x} = 0 $

Thus, $ 0 = 0 $, confirming that this function is also a solution.

It’s worth noting that differential equations can sometimes be presented in implicit form, where $ y $ is not isolated. In such cases, implicit differentiation may be required to verify the solution. This situation is less common but important to recognize.

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Verifying Solutions of Differential Equations Example 1

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Verifying Solutions of Differential Equations Example 1 Video Summary

To demonstrate that the given solution, $4x^2 = 2y^3 + 4y$, satisfies the corresponding differential equation, we will employ implicit differentiation. This method is necessary because $y$ appears in multiple terms, making it impossible to isolate it directly.

We start by differentiating both sides of the equation with respect to $x$. The left side, $4x^2$, differentiates to $8x$. For the right side, we apply the chain rule: the derivative of $2y^3$ is $6y^2 \frac{dy}{dx}$ and the derivative of $4y$ is $4 \frac{dy}{dx}$. Thus, we have:

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

Which simplifies to:

\[8x = 6y^2 \frac{dy}{dx} + 4 \frac{dy}{dx}\]

Next, we can factor out $\frac{dy}{dx}$ from the right side:

\[8x = \frac{dy}{dx}(6y^2 + 4)\]

To isolate $\frac{dy}{dx}$, we divide both sides by $6y^2 + 4$:

\[\frac{dy}{dx} = \frac{8x}{6y^2 + 4}\]

Now that we have the derivative, we substitute it back into the differential equation to verify if it holds true. The equation we are checking is:

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

Substituting $\frac{dy}{dx}$ gives us:

\[3y^2 + 2 \left(\frac{8x}{6y^2 + 4}\right) = 4x\]

To simplify the left side, we can factor out a 2 from the term $2 \cdot 8x$ and the denominator $6y^2 + 4$:

\[3y^2 + \frac{16x}{6y^2 + 4} = 4x\]

Now, we can multiply through by $6y^2 + 4$ to eliminate the fraction:

\[(3y^2)(6y^2 + 4) + 16x = 4x(6y^2 + 4)\]

Expanding both sides leads to:

\[18y^4 + 12y^2 + 16x = 24xy^2 + 16x\]

After canceling $16x$ from both sides, we are left with:

\[18y^4 + 12y^2 = 24xy^2\]

Rearranging confirms that both sides are equal, thus verifying that the original solution satisfies the differential equation. Therefore, we conclude that the given solution is indeed valid.

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Verifying Solutions of Differential Equations Example 2

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Verifying Solutions of Differential Equations Example 2 Video Summary

To verify that the function $ y(t) = -2 \cos(2t) $ is a solution to the initial value problem defined by the differential equation $ y'' + 4y = 0 $ with the initial condition $ y'(0) = 0 $, we first need to compute the first and second derivatives of the function.

The first derivative, denoted as $ y' $, is calculated as follows:

$ y' = \frac{d}{dt}(-2 \cos(2t)) = 4 \sin(2t) $

Next, we find the second derivative, $ y'' $:

$ y'' = \frac{d}{dt}(4 \sin(2t)) = 8 \cos(2t) $

Now that we have both the original function and its second derivative, we can substitute these into the differential equation:

$ y'' + 4y = 8 \cos(2t) + 4(-2 \cos(2t)) $

Substituting the values gives:

$ 8 \cos(2t) - 8 \cos(2t) = 0 $

This simplifies to:

$ 0 = 0 $

Since this statement is true, we have verified that the function satisfies the differential equation.

Next, we check the initial condition $ y'(0) = 0 $. Substituting $ t = 0 $ into the first derivative:

$ y'(0) = 4 \sin(2 \cdot 0) = 4 \sin(0) = 0 $

This confirms that the initial condition is satisfied. Therefore, the function $ y(t) = -2 \cos(2t) $ is indeed a solution to the initial value problem.

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Verifying Solutions of Differential Equations Example 3

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Verifying Solutions of Differential Equations Example 3 Video Summary

To determine if the function $ y = e^{4x} $ is a solution to the differential equation $ y'' - 4y' + 3y = 0 $, we first need to compute the first and second derivatives of $ y $.

The first derivative, denoted as $ y' $, is calculated using the chain rule:

$ y' = \frac{d}{dx}(e^{4x}) = 4e^{4x} $.

Next, we find the second derivative, $ y'' $, again applying the chain rule:

$ y'' = \frac{d}{dx}(4e^{4x}) = 16e^{4x} $.

Now that we have $ y $, $ y' $, and $ y'' $, we can substitute these into the differential equation:

$ y'' - 4y' + 3y = 16e^{4x} - 4(4e^{4x}) + 3(e^{4x}) $.

Expanding this gives:

$ 16e^{4x} - 16e^{4x} + 3e^{4x} $.

Combining the terms results in:

$ 0 + 3e^{4x} = 3e^{4x} $.

Setting this equal to zero, we have:

$ 3e^{4x} \neq 0 $.

Since $ 3e^{4x} $ does not equal zero, we conclude that $ y = e^{4x} $ is not a solution to the differential equation $ y'' - 4y' + 3y = 0 $.

This process illustrates the importance of verifying potential solutions to differential equations, as the phrasing of the problem can indicate whether a solution is expected or if it may not hold true.

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Verifying Solutions of Differential Equations Example 4

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Verifying Solutions of Differential Equations Example 4 Video Summary

To verify that the function $ y = \frac{c}{x^2} $ is a general solution to the differential equation $ x y' = -2y $, we first need to compute the derivative $ y' $. Recognizing that $ y $ can be rewritten as $ c x^{-2} $, we apply the power rule for differentiation. The power rule states that if $ y = kx^n $, then $ y' = knx^{n-1} $. Here, $ k = c $ and $ n = -2 $.

Using the power rule, we find:

$ y' = -2c x^{-3} = \frac{-2c}{x^3} $.

Next, we substitute $ y $ and $ y' $ into the differential equation. The left-hand side becomes:

$ x y' = x \left( \frac{-2c}{x^3} \right) = \frac{-2c}{x^2} $.

For the right-hand side, substituting $ y $ gives:

$ -2y = -2 \left( \frac{c}{x^2} \right) = \frac{-2c}{x^2} $.

Now, we compare both sides of the equation:

$ \frac{-2c}{x^2} = \frac{-2c}{x^2} $.

Since both sides are equal, we confirm that the function $ y = \frac{c}{x^2} $ satisfies the differential equation $ x y' = -2y $. Therefore, it is indeed a general solution to the given differential equation.

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Verifying Solutions of Differential Equations Example 5

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Verifying Solutions of Differential Equations Example 5 Video Summary

To verify that a given function is a solution to a differential equation, we start with the function y(t) = c e^{-6t}, where c is an arbitrary constant. The differential equation we need to check is y'(t) + 6y(t) = 0.

First, we find the derivative of the function. Using the chain rule, the derivative y'(t) is calculated as follows:

y'(t) = -6c e^{-6t}

Next, we substitute both y(t) and y'(t) into the differential equation:

-6c e^{-6t} + 6(c e^{-6t}) = 0

Now, simplifying the left-hand side:

-6c e^{-6t} + 6c e^{-6t} = 0

Both terms cancel each other out, resulting in:

0 = 0

This confirms that the equation holds true, verifying that y(t) = c e^{-6t} is indeed a solution to the differential equation y'(t) + 6y(t) = 0.

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Verifying Solutions of Differential Equations Example 6

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Verifying Solutions of Differential Equations Example 6 Video Summary

To illustrate this process, consider the function $ y = e^{2x} $ and the differential equation $ 3y' - 5y = e^{2x} $. First, we need to find the first derivative of the function, which is calculated using the chain rule: $ y' = 2e^{2x} $. Substituting $ y $ and $ y' $ into the differential equation gives:

\[3(2e^{2x}) - 5(e^{2x}) = e^{2x}\]

Performing the algebra on the left side results in:

\[6e^{2x} - 5e^{2x} = e^{2x}\]

This simplifies to $ e^{2x} = e^{2x} $, confirming that the function is indeed a solution.

Next, let's verify another example with the function $ y = 4 + \ln(x) $ and the differential equation $ x y'' + y' = 0 $. We first find the first and second derivatives. The first derivative is $ y' = \frac{1}{x} $, and the second derivative, using the power rule, is $ y'' = -\frac{1}{x^2} $. Substituting these into the differential equation yields:

\[x\left(-\frac{1}{x^2}\right) + \frac{1}{x} = 0\]

After simplifying, we have:

\[-\frac{1}{x} + \frac{1}{x} = 0\]

This results in $ 0 = 0 $, verifying that this function is also a solution.

In some cases, differential equations may be presented in implicit form, meaning $ y $ is not isolated. In such instances, implicit differentiation is required to verify the solution. While this situation is less common, it is essential to be prepared for it in your studies.

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Solutions to Basic Differential Equations

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Solutions to Basic Differential Equations Video Summary

To solve a basic differential equation, we start by understanding the relationship between a function and its derivative. In this case, we are given the differential equation $ y' = 4x^3 $ and need to find the particular solution that passes through the point (1, 4).

The first step is to find the general solution by integrating the right-hand side of the equation. We can do this by calculating the antiderivative:

\[ y = \int 4x^3 \, dx \]

Using the power rule for integration, we add one to the exponent and divide by the new exponent:

\[ y = x^4 + C \]

Here, $ C $ is the constant of integration, which represents a family of functions known as the general solution. Each value of $ C $ corresponds to a different curve on the graph, illustrating the various possible solutions to the differential equation.

To find the particular solution, we need to apply the initial condition provided, which is the point (1, 4). We substitute $ x = 1 $ and $ y = 4 $ into the general solution:

\[ 4 = 1^4 + C \]

Solving for $ C $, we find:

\[ 4 = 1 + C \]

\[ C = 4 - 1 = 3 \]

Now that we have determined $ C $, we can substitute it back into the general solution to obtain the particular solution:

\[ y = x^4 + 3 \]

This equation represents the specific curve that passes through the point (1, 4). In summary, to solve a basic differential equation, we first integrate to find the general solution and then apply an initial condition to determine the constant of integration, resulting in the particular solution.

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Solutions to Basic Differential Equations Example 7

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Solutions to Basic Differential Equations Example 7 Video Summary

To verify that the function $ y = c e^{x^2} $ is a solution to the differential equation $ y' = 2xy $, we first need to compute the derivative $ y' $. Using the chain rule, we find:

$ y' = c e^{x^2} \cdot \frac{d}{dx}(x^2) = c e^{x^2} \cdot 2x = 2c x e^{x^2} $.

Now, substituting $ y $ and $ y' $ into the differential equation, we have:

$ 2c x e^{x^2} = 2x(c e^{x^2}) $.

Both sides of the equation are equal, confirming that $ y = c e^{x^2} $ is indeed a solution to the differential equation.

Next, we need to find the particular solution that passes through the point $ (0, 6) $. We substitute $ x = 0 $ and $ y = 6 $ into the general solution:

$ 6 = c e^{0^2} $.

Since $ e^{0} = 1 $, this simplifies to:

$ 6 = c $.

Thus, the constant $ c $ is equal to 6. Therefore, the particular solution is:

$ y = 6 e^{x^2} $.

This function represents the specific solution that passes through the point $ (0, 6) $.

Problem

Find the particular solution to the differential equation $y^{'} = 2 e^{t} + 4 t$ given the initial condition $y of 0 equals 1$ .

$y = 2 e^{t} + 2 t^{2}$

$y = e^{t} + 2 t^{2} - 1$

$y = e^{t} + 2 t^{2} + 1$

$y = 2 e^{t} + 2 t^{2} - 1$

Problem

Find the general solution to the differential equation $\frac{d y}{d x} = - 2 x + 5 x^{2}$ .

$y = - x^{2} + \frac{5}{3} x^{3} + C$

$y = - x^{2} + 10 x + C$

$y = - x + 5 x + C$

$y = - x^{} + \frac{5}{3} x^{2} + C$

Problem

Find the particular solution to the differential equation $y^{'} = 2 \sin x + 3 \cos x$ given the initial condition $y of 0 equals 4$ .

$y = - 2 \cos x + 3 \sin x + 6$

$y = - 2 \cos x + 3 \sin x + 4$

$y = - 2 \cos x + 3 \sin x + 1$

$y = - 2 \cos x + 3 \sin x$

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Here’s what students ask on this topic:

What is the order of a differential equation, and how is it determined?

The order of a differential equation is determined by the highest derivative present in the equation. For example, if a differential equation contains both the first derivative (dy/dx) and the second derivative (d²y/dx²), the order is determined by the second derivative, making it a second-order differential equation. The order is crucial because it helps classify the equation and determine the methods used to solve it. Higher-order differential equations often require more complex techniques for finding solutions compared to first-order equations.

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How can you classify a differential equation as linear or nonlinear?

A differential equation is classified as linear if it satisfies three criteria: (1) the dependent variable and its derivatives are not multiplied by each other, (2) they are raised only to the first power, and (3) they are not inside another function (e.g., as arguments of trigonometric or exponential functions). If any of these conditions are violated, the equation is nonlinear. For example, y'' + 4y' - 3y = sin(e^x) is linear, while y'' - e^y = 0 is nonlinear because y appears as the argument of an exponential function.

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What does it mean to verify a solution to a differential equation?

To verify a solution to a differential equation, you substitute the given function and its derivatives into the equation to check if it satisfies the equation. For example, if the differential equation is 3y' - 5y = e^2x and the solution is y = e^2x, you calculate y' = 2e^2x and substitute both y and y' into the equation. Simplify the left-hand side and confirm that it equals the right-hand side. If the equation holds true, the solution is verified.

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How do you find the particular solution of a differential equation?

To find the particular solution of a differential equation, start by integrating the equation to find the general solution, which includes an arbitrary constant (C). Then, use the initial condition (a specific point on the curve) to solve for C. For example, if y' = 4x³ and the initial condition is y(1) = 4, integrate to get y = x⁴ + C. Substitute x = 1 and y = 4 to find C = 3. The particular solution is y = x⁴ + 3.

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What is the difference between a general solution and a particular solution in differential equations?

The general solution of a differential equation includes an arbitrary constant (C) and represents a family of solutions. It is obtained by integrating the equation without applying any initial conditions. The particular solution is a specific function derived from the general solution by using initial conditions to determine the value of C. For example, the general solution of y' = 4x³ is y = x⁴ + C, while the particular solution, given y(1) = 4, is y = x⁴ + 3.

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Basics of Differential Equations: Videos & Practice Problems

Classifying Differential Equations

Classifying Differential Equations Video Summary

State the order of the differential equation and indicate if it is linear or nonlinear. y′′′+3xy=4xy^{^{\prime\prime\prime}}+3xy=4\sqrt{x}

State the order of the differential equation and indicate if it is linear or nonlinear. d2ydx2−(dydx)(1−x)=2\frac{d^2y}{dx^2}-\left(\frac{dy}{dx}\right)(1-x)=2

State the order of the differential equation and indicate if it is linear or nonlinear. (y′′)2+6ety′=4t\left(y^{\prime\prime}\right)^2+6e^{t}y^{\prime}=4t

State the order of the differential equation and indicate if it is linear or nonlinear. y′∙y=3ety^{\prime}\bullet y=3e^{t}

Verifying Solutions of Differential Equations

Verifying Solutions of Differential Equations Video Summary

Verifying Solutions of Differential Equations Example 1

Verifying Solutions of Differential Equations Example 1 Video Summary

Verifying Solutions of Differential Equations Example 2

Verifying Solutions of Differential Equations Example 2 Video Summary

Verifying Solutions of Differential Equations Example 3

Verifying Solutions of Differential Equations Example 3 Video Summary

Verifying Solutions of Differential Equations Example 4

Verifying Solutions of Differential Equations Example 4 Video Summary

Verifying Solutions of Differential Equations Example 5

Verifying Solutions of Differential Equations Example 5 Video Summary

Verifying Solutions of Differential Equations Example 6

Verifying Solutions of Differential Equations Example 6 Video Summary

Solutions to Basic Differential Equations

Solutions to Basic Differential Equations Video Summary

Solutions to Basic Differential Equations Example 7

Solutions to Basic Differential Equations Example 7 Video Summary

Find the particular solution to the differential equation y′=2et+4ty^{\prime}=2e^{t}+4t given the initial condition y(0)=1y\left(0\right)=1.

Find the general solution to the differential equation dydx=−2x+5x2\frac{dy}{dx}=-2x+5x^2.

Find the particular solution to the differential equation y′=2sinx+3cosxy^{\prime}=2\sin x+3\cos x given the initial condition y(0)=4y\left(0\right)=4.

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Here’s what students ask on this topic:

What is the order of a differential equation, and how is it determined?

How can you classify a differential equation as linear or nonlinear?

What does it mean to verify a solution to a differential equation?

How do you find the particular solution of a differential equation?

What is the difference between a general solution and a particular solution in differential equations?

Your Calculus tutors

State the order of the differential equation and indicate if it is linear or nonlinear.
$y^{^{'''}} + 3 x y = 4 \sqrt{x}$

State the order of the differential equation and indicate if it is linear or nonlinear.
$\frac{d^{2} y}{d x^{2}} - (\frac{d y}{d x}) (1 - x) = 2$

State the order of the differential equation and indicate if it is linear or nonlinear.
${(y^{''})}^{2} + 6 e^{t} y^{'} = 4 t$

State the order of the differential equation and indicate if it is linear or nonlinear.
$y^{'} ∙ y = 3 e^{t}$

Find the particular solution to the differential equation $y^{'} = 2 e^{t} + 4 t$ given the initial condition $y of 0 equals 1$ .

Find the general solution to the differential equation $\frac{d y}{d x} = - 2 x + 5 x^{2}$ .

Find the particular solution to the differential equation $y^{'} = 2 \sin x + 3 \cos x$ given the initial condition $y of 0 equals 4$ .