First-Order Differential 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.

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

Answer

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 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 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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Question

Find the solution for the following differential equation:

where .

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?

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.

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

Answer

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 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 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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Question

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.

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

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

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

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

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

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

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

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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Question

Solve the differential equation

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

Answer

Rearranging the following equation

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

This satisfies the test of exactness, so integrating we have

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Question

Solve the given differential equation by separation of variables.

$x\frac{dy}{dx}=6y$

Answer

To solve this differential equation use separation of variables. This means move all terms containing to one side of the equation and all terms containing to the other side.

$x\frac{dy}{dx}=6y$

First, multiply each side by .

Now divide by on both sides.

$\frac{xdy}{y}=6dx$

Next, divide by on both sides.

$\frac{dy}{y}=\frac{6dx}{x}$

From here take the integral of both sides. Remember rules for logarithmic functions as they will be used in this problem.

$\\int\frac{dy}{y}=\int\frac{6dx}{x} \\\ln y=6\ln x \\e^{\ln y}=e^{6\ln x} \\y=e^{\ln x^6} \\y=cx^6$

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Question

Solve the following differential equation

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

Answer

So this is a separable differential equation. The first step is to move all of the x terms (including dx) to one side, and all of the y terms (including dy) to the other side.

So the differential equation we are given is: $\frac{dy}{dx}=3x^2y$

Which rearranged looks like: $\frac{dy}{y}=3x^2dx$

At this point, in order to solve for y, we need to take the anti-derivative of both sides: $\int \frac{dy}{y}=\int 3x^2dx$

Which equals:

And since this an anti-derivative with no bounds, we need to include the general constant C

So, solving for y, we raise e to the power of both sides:

$e^{ln(y)}=e^{x^3+C}$ which, simplified gives us our answer:

$y=Ce^{x^3}$

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Question

Solve the following separable differential equation: with .

Answer

The simplest way to solve a separable differential equation is to rewrite as $\frac{\mathrm{d} y}{\mathrm{d} t}$ and, by an abuse of notation, to "multiply both sides by dt". This yields

.

Next, we get all the y terms with dy and all the t terms with dt and integrate. Thus,

$\int \frac{dy}{y} = \int 5t^4dt$

Combining the constants of integration and exponentiating, we have

$|y| = e^{t^5 + C}$

$y = \pm e^Ce^{t^5}$

The plus minus and the can be combined into another arbitrary constant, yielding $y = Ce^{t^5}$ .

Plugging in our initial condition, we have

and

$y = 3e^{t^5}$

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Question

Is the following differential equation separable? If So, how does the equation separate?

$\frac{dy}{dt}= e^{t+y}$

Answer

Using exponential rules, we note that $e^{t+y}$ becomes . Meaning that the

differential equation is equivalent to:

$\frac{dy}{dt} = e^t e^y$

which by separation of variables is:

$\frac{dy}{e^y}= e^tdt$

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Question

Is the following differential equation separable, if so, how does the equation separate?

Answer

The differential equation cannot be written as and is therefore not separable.

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Question

Solve the general solution for the ODE:

$\frac{dy}{dx}=x^8y$

Answer

First the differential equation can be separated to:

$\frac{dy}{y} = x^8dx$

And then integrated simply to:

$\ln(y) = \frac{x^9}{9}+C$

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Question

Solve the given differential equation by separation of variables.

$x\frac{dy}{dx}=6y$

Answer

To solve this differential equation use separation of variables. This means move all terms containing to one side of the equation and all terms containing to the other side.

$x\frac{dy}{dx}=6y$

First, multiply each side by .

Now divide by on both sides.

$\frac{xdy}{y}=6dx$

Next, divide by on both sides.

$\frac{dy}{y}=\frac{6dx}{x}$

From here take the integral of both sides. Remember rules for logarithmic functions as they will be used in this problem.

$\\int\frac{dy}{y}=\int\frac{6dx}{x} \\\ln y=6\ln x \\e^{\ln y}=e^{6\ln x} \\y=e^{\ln x^6} \\y=cx^6$

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Question

Solve the following initial value problem: $6y' = y^4\sin(t)$ , .

Answer

This is a separable differential equation. The simplest way to solve this is to first rewrite as $\frac{\mathrm{d} y}{\mathrm{d} t}$ and then by an abuse of notation to "multiply both sides by dt." This yields $6dy = y^4\sin(t)dt$ . Then group all the y terms with dy and integrate, getting us to $\int \frac{6}{y^{4}}dy = \int \sin(t)dt = -\frac{2}{y^3} = -\cos(t) + C$ . Solving for y, we have $y = \left(\frac{\cos(t)}{2} + C\right)^{-1/3}$ . Plugging in our condition, we find $1 = \left( \frac{1}{2} + C\right )^{-1/3}$ . Raising both sides to the power of -1/3, we see $1 = \frac{1}{2} + C; C = \frac{1}{2}$ . Thus, our final solution is $y = \left(\frac{1 + \cos(t)}{2}\right)^{-1/3}$

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Question

Solve the following equation

$\frac{1}{2}\frac{dy}{dx} = \sqrt{y+1}\cos(x) \ \ \ y(\pi) = 0$

Answer

This is a separable ODE, so rearranging

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

Integrating

$2(y+1)^{1/2} = 2\sin{x} + C$

Plugging in the initial condition and solving gives us

Solving for gives us

$2(y+1)^{1/2} = 2\sin{x} + 2$

$(y+1) = (\sin{x} + 1)^2$

$y = (\sin{x} + 1)^2 - 1$

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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.

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

Answer

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 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 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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Question

Find the solution for the following differential equation:

where .

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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