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

Calculatrice Equation différentielle homogène

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

Go!
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 homogeneous differential equation. This solution was automatically generated by our smart calculator:

$\frac{dy}{dx}=-\frac{4x+3y}{2x+y}$
2

We can identify that the differential equation $\frac{dy}{dx}=\frac{-\left(4x+3y\right)}{2x+y}$ is homogeneous, since it is written in the standard form $\frac{dy}{dx}=\frac{M(x,y)}{N(x,y)}$, where $M(x,y)$ and $N(x,y)$ are the partial derivatives of a two-variable function $f(x,y)$ and both are homogeneous functions of the same degree

$\frac{dy}{dx}=\frac{-\left(4x+3y\right)}{2x+y}$
3

Use the substitution: $y=ux$

$\frac{u\cdot dx+x\cdot du}{dx}=\frac{-\left(4x+3ux\right)}{2x+ux}$

Factor the polynomial $2x+ux$ by it's greatest common factor (GCF): $x$

$\frac{u\cdot dx+x\cdot du}{dx}=\frac{-\left(4x+3ux\right)}{x\left(2+u\right)}$

Expand the fraction $\frac{u\cdot dx+x\cdot du}{dx}$ into $2$ simpler fractions with common denominator $dx$

$\frac{u\cdot dx}{dx}+\frac{x\cdot du}{dx}=\frac{-\left(4x+3ux\right)}{x\left(2+u\right)}$

Simplify the resulting fractions

$u+\frac{x\cdot du}{dx}=\frac{-\left(4x+3ux\right)}{x\left(2+u\right)}$

Multiply the single term $-1$ by each term of the polynomial $\left(4x+3ux\right)$

$u+\frac{x\cdot du}{dx}=\frac{-4x-3ux}{x\left(2+u\right)}$

Factor the polynomial $-4x-3ux$ by it's greatest common factor (GCF): $-x$

$u+\frac{x\cdot du}{dx}=\frac{-x\left(4+3u\right)}{x\left(2+u\right)}$

Simplify the fraction $\frac{-x\left(4+3u\right)}{x\left(2+u\right)}$ by $x$

$u+\frac{x\cdot du}{dx}=\frac{-\left(4+3u\right)}{2+u}$

Multiply the single term $-1$ by each term of the polynomial $\left(4+3u\right)$

$u+\frac{x\cdot du}{dx}=\frac{-4-3u}{2+u}$

We need to isolate the dependent variable $u$, we can do that by simultaneously subtracting $u$ from both sides of the equation

$\frac{x\cdot du}{dx}=\frac{-4-3u}{2+u}-u$

Combine all terms into a single fraction with $2+u$ as common denominator

$\frac{x\cdot du}{dx}=\frac{-4-3u-2u-u^2}{2+u}$

Combining like terms $-3u$ and $-2u$

$\frac{x\cdot du}{dx}=\frac{-4-5u-u^2}{2+u}$

Group the terms of the differential equation. Move the terms of the $u$ variable to the left side, and the terms of the $x$ variable to the right side of the equality

$\frac{2+u}{-4-5u-u^2}du=\frac{1}{x}dx$

Simplify the expression $\frac{2+u}{-4-5u-u^2}du$

$\frac{2+u}{-\left(u+1\right)\left(u+4\right)}du=\frac{1}{x}dx$
4

Expand and simplify

$\frac{2+u}{-\left(u+1\right)\left(u+4\right)}du=\frac{1}{x}dx$
5

Integrate both sides of the differential equation, the left side with respect to $u$, and the right side with respect to $x$

$\int\frac{2+u}{-\left(u+1\right)\left(u+4\right)}du=\int\frac{1}{x}dx$

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

$-\int\frac{2+u}{\left(u+1\right)\left(u+4\right)}du$

Rewrite the fraction $\frac{2+u}{\left(u+1\right)\left(u+4\right)}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{1}{3\left(u+1\right)}+\frac{2}{3\left(u+4\right)}$

Expand the integral $\int\left(\frac{1}{3\left(u+1\right)}+\frac{2}{3\left(u+4\right)}\right)du$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$-\int\frac{1}{3\left(u+1\right)}du-\int\frac{2}{3\left(u+4\right)}du$

Take the constant $\frac{1}{3}$ out of the integral

$- \left(\frac{1}{3}\right)\int\frac{1}{u+1}du-\int\frac{2}{3\left(u+4\right)}du$

Multiply the fraction and term in $- \left(\frac{1}{3}\right)\int\frac{1}{u+1}du$

$-\frac{1}{3}\int\frac{1}{u+1}du-\int\frac{2}{3\left(u+4\right)}du$

Take the constant $\frac{1}{3}$ out of the integral

$-\frac{1}{3}\int\frac{1}{u+1}du- \left(\frac{1}{3}\right)\int\frac{2}{u+4}du$

Multiply the fraction and term in $- \left(\frac{1}{3}\right)\int\frac{2}{u+4}du$

$-\frac{1}{3}\int\frac{1}{u+1}du-\frac{1}{3}\int\frac{2}{u+4}du$

Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=1$, $x=u$ and $n=1$

$1\left(-\frac{1}{3}\right)\ln\left|u+1\right|-\frac{1}{3}\int\frac{2}{u+4}du$

Any expression multiplied by $1$ is equal to itself

$-\frac{1}{3}\ln\left|u+1\right|-\frac{1}{3}\int\frac{2}{u+4}du$

Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=4$, $x=u$ and $n=2$

$-\frac{1}{3}\ln\left|u+1\right|+2\left(-\frac{1}{3}\right)\ln\left|u+4\right|$

Multiply the fraction and term in $2\left(-\frac{1}{3}\right)\ln\left|u+4\right|$

$-\frac{1}{3}\ln\left|u+1\right|-\frac{2}{3}\ln\left|u+4\right|$
6

Solve the integral $\int\frac{2+u}{-\left(u+1\right)\left(u+4\right)}du$ and replace the result in the differential equation

$-\frac{1}{3}\ln\left|u+1\right|-\frac{2}{3}\ln\left|u+4\right|=\int\frac{1}{x}dx$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\ln\left|x\right|$

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

$\ln\left|x\right|+C_0$
7

Solve the integral $\int\frac{1}{x}dx$ and replace the result in the differential equation

$-\frac{1}{3}\ln\left|u+1\right|-\frac{2}{3}\ln\left|u+4\right|=\ln\left|x\right|+C_0$
8

Replace $u$ with the value $\frac{y}{x}$

$-\frac{1}{3}\ln\left(\frac{y}{x}+1\right)-\frac{2}{3}\ln\left(\frac{y}{x}+4\right)=\ln\left(x\right)+C_0$

Réponse finale au problème

$-\frac{1}{3}\ln\left(\frac{y}{x}+1\right)-\frac{2}{3}\ln\left(\frac{y}{x}+4\right)=\ln\left(x\right)+C_0$

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 !