👉 Essayez maintenant NerdPal! Notre nouvelle application de mathématiques sur iOS et Android
  1. calculatrices
  2. Équation Différentielle Linéaire

Calculatrice Équation différentielle linéaire

Résolvez vos problèmes de mathématiques avec notre calculatrice Équation différentielle linéaire étape par étape. Améliorez vos compétences en mathématiques grâce à notre longue liste de problèmes difficiles. Retrouvez tous nos calculateurs ici.

Mode symbolique
Mode texte
Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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 $y$

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

Réponse finale au problème

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

Vous avez des difficultés en mathématiques ?

Accédez à des solutions détaillées, étape par étape, à des milliers de problèmes, dont le nombre augmente chaque jour !