Linear & Exact Equations - Differential Equations

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Question

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

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Answer

First, divide by on both sides of the equation.

Identify the factor term.

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

Integrate the factor.

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 to get the general solution.

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 .

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Question

Find the solution for the following differential equation:

where .

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Answer

This equation can be put into the form as follows:

. Differential equations in this form can be solved by use of integrating factor. To solve, take 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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Question

Consider the differential equation

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

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Answer

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

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Question

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

Tap to reveal answer

Answer

First, divide by on both sides of the equation.

Identify the factor term.

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

Integrate the factor.

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 to get the general solution.

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

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Question

Is the following differential equation exact?

If so, find the general solution.

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Answer

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 and . The second condition is that . Taking the partial derivatives, we find that and . 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 and thus that and .

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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Question

Is the following differential equation exact? If so, find the general solution.

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Answer

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 and . The second condition is that . Taking the partial derivatives, we find that and . As these are unequal, we do not have an exact equation.

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Question

Solve the Following Equation

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

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Answer

Since this is in the form of a linear equation

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 we get

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

Integrating

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

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

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

Our solution is

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Question

Find the general solution of the differential equation

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

Tap to reveal answer

Answer

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

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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Question

Solve the differential equation

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

Tap to reveal answer

Answer

Rearranging the following equation

This satisfies the test of exactness, so integrating we have

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Question

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

Tap to reveal answer

Answer

First, divide by on both sides of the equation.

Identify the factor term.

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

Integrate the factor.

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 to get the general solution.

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 .

← Didn't Know|Knew It →

Question

Find the solution for the following differential equation:

where .

Tap to reveal answer

Answer

This equation can be put into the form as follows:

. Differential equations in this form can be solved by use of integrating factor. To solve, take 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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Question

Consider the differential equation

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

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

First, divide by on both sides of the equation.

Identify the factor term.

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

Integrate the factor.

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 to get the general solution.

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

← Didn't Know|Knew It →

Question

Is the following differential equation exact?

If so, find the general solution.

Tap to reveal answer

Answer

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 and . The second condition is that . Taking the partial derivatives, we find that and . 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 and thus that and .

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.

← Didn't Know|Knew It →

Question

Is the following differential equation exact? If so, find the general solution.

Tap to reveal answer

Answer

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 and . The second condition is that . Taking the partial derivatives, we find that and . As these are unequal, we do not have an exact equation.

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Question

Solve the Following Equation

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

Tap to reveal answer

Answer

Since this is in the form of a linear equation

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 we get

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

Integrating

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

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

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

Our solution is

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Question

Find the general solution of the differential equation

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

Tap to reveal answer

Answer

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

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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Question

Solve the differential equation

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

Tap to reveal answer

Answer

Rearranging the following equation

This satisfies the test of exactness, so integrating we have

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

First, divide by on both sides of the equation.

Identify the factor term.

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

Integrate the factor.

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 to get the general solution.

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 .

← Didn't Know|Knew It →

Question

Find the solution for the following differential equation:

where .

Tap to reveal answer

Answer

This equation can be put into the form as follows:

. Differential equations in this form can be solved by use of integrating factor. To solve, take 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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