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 d x 2 d 2 y − ( d x d y ) ( 1 − x ) = 2

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

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

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

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

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

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

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

1 s t 1^{st} 1 s t order, linear

1 s t 1^{st} 1 s t order, nonlinear

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

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

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

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

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

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

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

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. 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. For example, the general solution of $y = x^{4} + c$ can be specified using a point like (1, 4) to determine the constant $c$ .

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 for 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 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 only raised to the first power, and are not inside any function, this equation is classified as a 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} $, classifying it as 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 also nonlinear.

As we continue to explore differential equations, understanding these classifications will aid in selecting appropriate methods for solving them. Mastery of these concepts is essential for progressing in the study of differential equations.

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

$1^{st}$ order; linear

$3^{rd}$ order; linear

$1^{st}$ order; nonlinear

Problem

State the order of the differential equation and indicate if it is linear or nonlinear.

$\frac{d^2y}{dx^2}-\left(\frac{dy}{dx}\right)(1-x)=2$

$2^{nd}$ order, linear

$2^{nd}$ order, nonlinear

$3^{rd}$ order, nonlinear

$3^{rd}$ order, linear

Problem

State the order of the differential equation and indicate if it is linear or nonlinear.

$\left(y^{\prime\prime}\right)^2+6e^{t}y^{\prime}=4t$

$2^{nd}$ order, linear

$2^{nd}$ order, nonlinear

$1^{st}$ order, linear

$1^{st}$ order, nonlinear

Problem

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

$1^{st}$ order, linear

$1^{st}$ order, nonlinear

$2^{nd}$ 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 against the equation.

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:

$ y' = 2e^{2x} $

Next, we substitute $ y $ and $ y' $ into the differential equation:

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

Performing the algebra gives:

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

Which simplifies to:

$ e^{2x} = e^{2x} $

This confirms that the function is indeed a solution.

Now, 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:

$ y' = \frac{1}{x} $

$ 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 \

Again, this confirms that the function is a valid 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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Verifying Solutions of Differential Equations Example 1

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Verifying Solutions of Differential Equations Example 1 Video Summary

To demonstrate that the 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. We replace $y'$ with $\frac{8x}{6y^2 + 4}$ in the equation:

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

To simplify the left side, we can factor out a 2 from the term $8x$ 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:

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

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

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

Since both sides are equal, we confirm that the original solution indeed satisfies the differential equation. This process illustrates the effectiveness of implicit differentiation in handling equations where $y$ cannot be easily isolated.

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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 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}) $.

Calculating each term gives:

First term: $ 16e^{4x} $
Second term: $ -16e^{4x} $ (since $ 4 \times 4 = 16 $)
Third term: $ 3e^{4x} $

Combining these results, we have:

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

Setting this equal to zero, we find:

$ 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 in differential equations, as the phrasing of the problem may indicate that a solution is not guaranteed.

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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 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 of the equation 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 4

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Verifying Solutions of Differential Equations Example 4 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) = \frac{d}{dt}(C e^{-6t}) = -6C e^{-6t} $.

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

$ y'(t) + 6y(t) = -6C e^{-6t} + 6(C e^{-6t}) $.

Now, simplifying the left-hand side:

$ -6C e^{-6t} + 6C e^{-6t} = 0 $.

This results in:

$ 0 = 0 $.

Since both sides of the equation are equal, we have successfully verified that the function $ 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 5

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Verifying Solutions of Differential Equations Example 5 Video Summary

$ y' = 2e^{2x} $

Next, we substitute $ y $ and $ y' $ into the differential equation:

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

Carrying out the algebra gives:

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

Which simplifies to:

$ e^{2x} = e^{2x} $

This confirms 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:

$ y' = \frac{1}{x} $

$ 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 find:

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

This results in the true statement $ 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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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. Since $ y' $ represents the derivative of $ y $, we can find $ y $ 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:

\[ y = \frac{4}{4} x^{4} + C = x^{4} + C \]

This expression represents the general solution of the differential equation, where $ C $ is the constant of integration. The general solution encompasses a family of functions, each corresponding to different values of $ C $.

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:

\[ y = x^{4} + 3 \]

This equation represents the particular solution to the differential equation that passes through the point (1, 4). The graph of this function shows the specific curve corresponding to this initial condition, while other curves represent different solutions based on varying values of $ C $.

In summary, to solve a basic differential equation, integrate to find the general solution, then apply an initial condition to determine the specific constant, resulting in the particular solution.

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Solutions to Basic Differential Equations Example 6

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Solutions to Basic Differential Equations Example 6 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, the derivative of $ y $ is given by:

$ 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) $. To do this, 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. We can now express the particular solution as:

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

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

Problem

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

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

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

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

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

Problem

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

$y=-x^2+\frac53x^3+C$

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

$y = - x + 5 x + C$

$y=-x^{}+\frac53x^2+C$

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Here’s what students ask on this topic:

What is the difference between linear and nonlinear differential equations?

A linear differential equation is one where the dependent variable (e.g., y) and its derivatives appear only to the first power, are not multiplied by each other, and are not inside other functions. For example, $y^{''} + 3 y^{'} - 2 y = sin(x)$ is linear. In contrast, a nonlinear differential equation violates one or more of these conditions. For instance, $y'^{2} + e y = 0$ is nonlinear because $y'^{2}$ involves a power greater than one, and $e y$ places y inside a function.

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How do you determine the order of a differential equation?

The order of a differential equation is determined by the highest derivative of the dependent variable present in the equation. For example, in $y^{''} + 4 y^{'} - 3 y = sin(x)$ , the highest derivative is $y^{''}$ (the second derivative), so it is a second-order differential equation. Similarly, if the equation contains $y^{''''}$ (the fourth derivative), it would be a fourth-order differential equation.

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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 $3 y^{'} - 5 y = e x$ and the proposed solution is $y = e x$ , you calculate $y^{'}$ , substitute both $y$ and $y^{'}$ into the equation, and simplify. If both sides of the equation are equal, 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, first integrate the equation to find the general solution, which includes an arbitrary constant $c$ . Then, use the given initial condition (a specific point on the solution curve) to solve for $c$ . For example, if $y^{'} = 4 x^{3}$ and the initial condition is $y (1) = 4$ , integrate to get $y = x^{4} + c$ . Substitute $x = 1$ and $y = 4$ to solve for $c$ , then write the particular solution.

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What is the general solution of a first-order differential equation?

The general solution of a first-order differential equation is the antiderivative of the given derivative, including an arbitrary constant of integration $c$ . For example, if $y^{'} = 2 x$ , the general solution is $y = x^{2} + c$ . This represents a family of functions, where each value of $c$ corresponds to a different curve.

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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)=2dx2d2y−(dxdy)(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(y′′)2+6ety′=4t

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

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

Solutions to Basic Differential Equations

Solutions to Basic Differential Equations Video Summary

Solutions to Basic Differential Equations Example 6

Solutions to Basic Differential Equations Example 6 Video Summary

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

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

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What is the difference between linear and nonlinear differential equations?

How do you determine the order of a differential equation?

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 general solution of a first-order differential equation?

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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.

$\frac{d^2y}{dx^2}-\left(\frac{dy}{dx}\right)(1-x)=2$

State the order of the differential equation and indicate if it is linear or nonlinear.

$\left(y^{\prime\prime}\right)^2+6e^{t}y^{\prime}=4t$

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

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

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