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Differential Equations : Linear & Exact Equations

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Example Questions

Example Question #1 : Linear & Exact Equations

Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

$x\frac{dy}{dx}+3y=x^4-x$

Possible Answers:

$\\y=\frac{1}{7}x^3-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x^3+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x^2+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

Correct answer:

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

Explanation:

First, divide by on both sides of the equation.

$\\\frac{dy}{dx}+\frac{3y}{x}=\frac{x^4}{x}-\frac{x}{x} \\\\\frac{dy}{dx}+\frac{3y}{x}=x^3-1$

Identify the factor p(x) term.

$p(x)=\frac{3}{x}$

Integrate the factor.

$e^{\int p(x)}=e^{\frac{3}{x}}=e^{3\ln x}=e^{\ln x^3}=x^3$

Substitute this value back in and integrate the equation.

$\\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\\\x^3y=\int x^6-x^3dx \\\\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c$

Now divide by x^3 to get the general solution.

$\\y=\frac{1}{7}\frac{x^7}{x^3}-\frac{1}{4}\frac{x^4}{x^3}+\frac{c}{x^3} \\\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3}$

The transient term means a term that when the values get larger the term itself gets smaller. Therefore the transient term for this function is $cx^{-3}$ .

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Example Question #1 : Linear & Exact Equations

Find the solution for the following differential equation:

y' = 3t^2y + t^2 where y(0) = 0 .

Possible Answers:

$y = -\frac{1}{3}$

$y = \frac{e^{t^3} - 1}{3}$

$y = \frac{e^{t^3}}{3}$

$y = e^{t^3}-\frac{1}{3}$

y = -t^3

Correct answer:

$y = \frac{e^{t^3} - 1}{3}$

Explanation:

This equation can be put into the form y' + p(t)y = q(t) as follows:

y' - 3t^2y = t^2 . Differential equations in this form can be solved by use of integrating factor. To solve, take p(t) = -3t^2 and solve for $\mu = e^{\int p(t)dt} = e^{\int -3t^2dt} = e^{-t^3}$

Note, when using integrating factors, the +C constant is irrelevant as we only need one solution, not infinitely many. Thus, we have set C to 0.

Next, note that $(\mu y)' = y'\mu + \mu'y = e^{-t^3}(y' - 3t^2y) = \mu q(t) = e^{-t^3}t^2$

Or more simply, $(\mu y)' = t^2e^{-t^3}$ . Integrating both sides using substitution of variables we find

$\mu y = \int t^2e^{-t^3}dt = -\frac{e^{-t^3}}{3} + C$

Finally dividing by $\mu = e^{-t^3}$ , we see

$y = -\frac{1}{3} + Ce^{t^3}$ . Plugging in our initial condition,

$0 = -\frac{1}{3} + C$

So $C = \frac{1}{3}$

And $y = \frac{e^{t^3} - 1}{3}$ .

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Example Question #11 : First Order Differential Equations

Consider the differential equation

y'+y^2=t^4e^t

Which of the terms in the differential equation make the equation nonlinear?

Possible Answers:

The y^2 term makes the differential equation nonlinear.

The t^4 term makes the differential equation nonlinear

The e^t term makes the differential equation nonlinear.

The term makes the differential equation nonlinear

Correct answer:

The y^2 term makes the differential equation nonlinear.

Explanation:

The term y^2 makes the differential equation nonlinear because a linear equation has the form of

y'+f(t)y=g(t)

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Example Question #2 : Linear & Exact Equations

Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

$x\frac{dy}{dx}+3y=x^4-x$

Possible Answers:

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{3}$

$\\y=\frac{1}{7}x^3-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x^3+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x^2+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

Correct answer:

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

Explanation:

First, divide by on both sides of the equation.

$\\\frac{dy}{dx}+\frac{3y}{x}=\frac{x^4}{x}-\frac{x}{x} \\\\\frac{dy}{dx}+\frac{3y}{x}=x^3-1$

Identify the factor p(x) term.

$p(x)=\frac{3}{x}$

Integrate the factor.

$e^{\int p(x)}=e^{\frac{3}{x}}=e^{3\ln x}=e^{\ln x^3}=x^3$

Substitute this value back in and integrate the equation.

$\\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\\\x^3y=\int x^6-x^3dx \\\\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c$

Now divide by x^3 to get the general solution.

$\\y=\frac{1}{7}\frac{x^7}{x^3}-\frac{1}{4}\frac{x^4}{x^3}+\frac{c}{x^3} \\\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3}$

The transient term means a term that when the values get larger the term itself gets smaller. Therefore the transient term for this function is $cx^{-3}$

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Example Question #1 : Linear & Exact Equations

Is the following differential equation exact? $2xy^2 + y^2 + 1 + (2x^2y + 2xy + 3)\frac{\mathrm{d} y}{\mathrm{d} x} = 0$

If so, find the general solution.

Possible Answers:

Yes. x^2y^2 + xy^2 + x + 3y = c

Yes. 2x^2y + 2xy^2 + y = c

No. The equation does not take the proper form.

No. The partial derivatives in the equation do not have the correct relationship.

Yes. x^2y + xy^2 + 3x + y = c

Correct answer:

Yes. x^2y^2 + xy^2 + x + 3y = c

Explanation:

For a differential equation to be exact, two things must be true. First, it must take the form $M(x,y) + N(x,y)\frac{\mathrm{d} y}{\mathrm{d} x} = 0$ . In our case, this is true, with M = 2xy^2 + y^2 + 1 and N = 2x^2y + 2xy + 3 . The second condition is that M_y = N_x . Taking the partial derivatives, we find that M_y =4xy + 2y and N_x = 4xy + 2y . As these are equal, we have an exact equation.

Next we find a $\Psi$ such that $\Psi_x = M$ and $\Psi_y = N$ . To do this, we can integrate with respect to or we can integrate with respect to Here, we choose arbitrarily to integrate .

$\Psi = \int \Psi_x dx = \int Mdx = \int(2xy^2 + y^2 + 1)dx =x^2y^2 + xy^2 + x + h(y)$

We aren't quite done yet, because when taking a multivariate integral, the constant of integration can now be a function of y instead of just a constant. However, we know that $\Psi_y = N$ , so taking the partial derivative, we find that 2x^2y + 2xy + h'(y) = 2x^2y + 2xy + 3 and thus that h'(y) = 3 and h(y)=3y .

We now know that $\Psi = x^2y^2 + xy^2 + x + 3y$ , and the point of finding psi was so that we could rewrite $M(x,y) + N(x,y)\frac{\mathrm{d} y}{\mathrm{d} x} = \Psi_x + \Psi_y\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} \Psi}{\mathrm{d} x} = 0$ , and because the derivative of psi is 0, we know it must have been a constant. Thus, our final answer is

$\Psi = x^2y^2 + xy^2 + x + 3y = c$ .

If you have an initial value, you can solve for c and have an implicit solution.

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Example Question #1 : Linear & Exact Equations

Is the following differential equation exact? $2x^2y + y + (x^2y + xy)\frac{\mathrm{d} y}{\mathrm{d} x} = 0$ If so, find the general solution.

Possible Answers:

No. The partial derivatives in the equation do not have the right relationship.

Yes. x^2y + xy^2 + x = c

No. The equation does not take the right form.

Yes. 4x^2y + 2xy = c

Yes. 4x^2y + 2xy^2 + y = c

Correct answer:

No. The partial derivatives in the equation do not have the right relationship.

Explanation:

For a differential equation to be exact, two things must be true. First, it must take the form $M(x,y) + N(x,y)\frac{\mathrm{d} y}{\mathrm{d} x} = 0$ . In our case, this is true, with M = 2xy^2 + y and N = x^2y + xy . The second condition is that M_y = N_x . Taking the partial derivatives, we find that M_y =4xy + 1 and N_x = 2xy + y . As these are unequal, we do not have an exact equation.

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Example Question #1 : Linear & Exact Equations

Solve the Following Equation

$y' - \frac{y}{x} = xe^x \ \ \ y(1) = e-1$

Possible Answers:

y = xe^x - 1

y = e^x + x

y = xe^x - x

y = e^x - x

y = xe^x + x

Correct answer:

y = xe^x - x

Explanation:

Since this is in the form of a linear equation

y' + P(x)y = Q(x)

we calculate the integration factor

$I(x) = e^{\int P(x)} = e^{\int-\frac{1}{x} \ dx} = e^{-\ln x} = \frac{1}{x}$

Multiplying by I(x) we get

$(y \times \frac{1}{x})' = e^x$

Integrating

$\frac{y}{x} = e^x + C$

y = xe^x + Cx

Plugging in the Initial Condition to solve for the Constant we get

$e - 1 = e + C \implies C = -1$

Our solution is

y = xe^x - x

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Example Question #12 : First Order Differential Equations

Find the general solution of the differential equation

$\frac{dy}{dx} - 5y = -\frac{5}{2}xy^3$

Possible Answers:

$y = \left( \frac{x}{2} + Ce^{-10x} \right)^{-1/2}$

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}$

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)$

None of the other answers

$y = \left( x - \frac{1}{10} + Ce^{-10x} \right)^{-1/2}$

Correct answer:

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}$

Explanation:

This is a Bernoulli Equation of the form

$\frac{dy}{dx} + P(x) = Q(x)y^n$

which requires a substitution

$v = y^{1-n}$

to transform it into a linear equation

Rearranging our equation gives us

$y^{-3} \frac{dy}{dx} - 5y^{-2} = -\frac{5}{2}x$

Substituting $v = y^{-2}$

$-\frac{1}{2} \frac{dv}{dx} - 5v = -\frac{5}{2}x$

$\frac{dv}{dx} + 10v = 5x$

Solving the linear ODE gives us

$v = \frac{x}{2} - \frac{1}{20} + Ce^{-10x}$

Substituting in $v = y^{-2}$ and solving for

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}$

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Example Question #3 : Linear & Exact Equations

Solve the differential equation

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

Possible Answers:

xy+x = C

None of the other answers

xy^2+x = C

x^2y^2+x = C

x^2y+x = C

Correct answer:

x^2y^2+x = C

Explanation:

Rearranging the following equation

$(2xy^2 + 1)\ dx +2x^2y \ dy = 0$

This satisfies the test of exactness, so integrating we have

x^2y^2+x = C

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Differential Equations : Linear & Exact Equations

Study concepts, example questions & explanations for Differential Equations

All Differential Equations Resources

Example Questions

Example Question #1 : Linear & Exact Equations

Example Question #1 : Linear & Exact Equations

Example Question #11 : First Order Differential Equations

Example Question #2 : Linear & Exact Equations

Example Question #1 : Linear & Exact Equations

Example Question #1 : Linear & Exact Equations

Example Question #1 : Linear & Exact Equations

Example Question #12 : First Order Differential Equations

Example Question #3 : Linear & Exact Equations

All Differential Equations Resources

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