First-Order Differential Equations - Differential Equations

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Question

Solve the differential equation

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

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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.

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

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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

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

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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

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

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

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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 .

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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.

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

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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.

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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? $2x^2y + y + (x^2y + xy)\frac{\mathrm{d} y}{\mathrm{d} x} = 0$ 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

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

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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 following separable differential equation: with .

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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?

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

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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?

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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:

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

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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.

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

Tap to reveal answer

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: $6y' = y^4\sin(t)$ , .

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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

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

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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.

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

Tap to reveal answer

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 .

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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