Linear Differential Equation Calculator & Solver

1

Here, we show you a step-by-step solved example of linear differential equation. This solution was automatically generated by our smart calculator:

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

2

We can identify that the differential equation has the form: $\frac{dy}{dx} + P(x)\cdot y(x) = Q(x)$, so we can classify it as a linear first order differential equation, where $P(x)=2$ and $Q(x)=x$. In order to solve the differential equation, the first step is to find the integrating factor $\mu(x)$

$\displaystyle\mu\left(x\right)=e^{\int P(x)dx}$

Compute the integral

$\int2dx$

The integral of a constant is equal to the constant times the integral's variable

$2x$

3

To find $\mu(x)$, we first need to calculate $\int P(x)dx$

$\int P(x)dx=\int2dx=2x$

4

So the integrating factor $\mu(x)$ is

$\mu(x)=e^{2x}$

5

Now, multiply all the terms in the differential equation by the integrating factor $\mu(x)$ and check if we can simplify

$\frac{dy}{dx}e^{2x}+2ye^{2x}=xe^{2x}$

6

We can recognize that the left side of the differential equation consists of the derivative of the product of $\mu(x)\cdot y(x)$

$\frac{d}{dx}\left(e^{2x}y\right)=xe^{2x}$

7

Integrate both sides of the differential equation with respect to $dx$

$\int\frac{d}{dx}\left(e^{2x}y\right)dx=\int xe^{2x}dx$

8

Simplify the left side of the differential equation

$e^{2x}y=\int xe^{2x}dx$

We can solve the integral $\int xe^{2x}dx$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$

First, identify or choose $u$ and calculate it's derivative, $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^{2x}dx}\\ \displaystyle{\int dv=\int e^{2x}dx}\end{matrix}$

Solve the integral to find $v$

$v=\int e^{2x}dx$

We can solve the integral $\int e^{2x}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=2x$

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by finding the derivative of the equation above

$du=2dx$

Isolate $dx$ in the previous equation

$dx=\frac{du}{2}$

Substituting $u$ and $dx$ in the integral and simplify

$\int\frac{e^u}{2}du$

Take the constant $\frac{1}{2}$ out of the integral

$\frac{1}{2}\int e^udu$

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$\frac{1}{2}e^u$

Replace $u$ with the value that we assigned to it in the beginning: $2x$

$\frac{1}{2}e^{2x}$

Now replace the values of $u$, $du$ and $v$ in the last formula

$\frac{1}{2}e^{2x}x-\frac{1}{2}\int e^{2x}dx$

We can solve the integral $\int e^{2x}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=2x$

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by finding the derivative of the equation above

$du=2dx$

Isolate $dx$ in the previous equation

$dx=\frac{du}{2}$

Substituting $u$ and $dx$ in the integral and simplify

$\frac{1}{2}e^{2x}x-\frac{1}{2}\int\frac{e^u}{2}du$

Take the constant $\frac{1}{2}$ out of the integral

$\frac{1}{2}e^{2x}x-\frac{1}{2}\cdot \frac{1}{2}\int e^udu$

Multiplying fractions $-\frac{1}{2} \times \frac{1}{2}$

$\frac{1}{2}e^{2x}x-\frac{1}{4}\int e^udu$

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$\frac{1}{2}e^{2x}x-\frac{1}{4}e^u$

Replace $u$ with the value that we assigned to it in the beginning: $2x$

$\frac{1}{2}e^{2x}x-\frac{1}{4}e^{2x}$

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\frac{1}{2}e^{2x}x-\frac{1}{4}e^{2x}+C_0$

9

Solve the integral $\int xe^{2x}dx$ and replace the result in the differential equation

$e^{2x}y=\frac{1}{2}e^{2x}x-\frac{1}{4}e^{2x}+C_0$

Multiplying the fraction by $e^{2x}x$

$e^{2x}y=\frac{1e^{2x}x}{2}-\frac{1}{4}e^{2x}+C_0$

Any expression multiplied by $1$ is equal to itself

$e^{2x}y=\frac{e^{2x}x}{2}-\frac{1}{4}e^{2x}+C_0$

Multiplying the fraction by $e^{2x}$

$e^{2x}y=\frac{e^{2x}x}{2}+\frac{-e^{2x}}{4}+C_0$

Multiplying the fraction by $e^{2x}x$

$e^{2x}y=\frac{1e^{2x}x}{2}-\frac{1}{4}e^{2x}+C_0$

Any expression multiplied by $1$ is equal to itself

$e^{2x}y=\frac{e^{2x}x}{2}-\frac{1}{4}e^{2x}+C_0$

Multiply the equation by the reciprocal of $e^{2x}$

$y=e^{-2x}\left(\frac{e^{2x}x}{2}+\frac{-e^{2x}}{4}+C_0\right)$

10

Find the explicit solution to the differential equation. We need to isolate the variable $y$

$y=e^{-2x}\left(\frac{e^{2x}x}{2}+\frac{-e^{2x}}{4}+C_0\right)$

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