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Calculatrice Equation différentielle homogène

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1

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

$\left(y^2+2xy\right)dx-x^2dy=0$
2

We can identify that the differential equation $\left(y^2+2xy\right)dx-x^2dy=0$ is homogeneous, since it is written in the standard form $M(x,y)dx+N(x,y)dy=0$, 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

$\left(y^2+2xy\right)dx-x^2dy=0$
3

Use the substitution: $y=ux$

$\left(\left(ux\right)^2+2xux\right)dx-x^2\left(u\cdot dx+x\cdot du\right)=0$

Multiply the single term $-x^2$ by each term of the polynomial $\left(u\cdot dx+x\cdot du\right)$

$\left(\left(ux\right)^2+2x^2u\right)dx-ux^2\cdot dx-x^{3}du=0$

The power of a product is equal to the product of it's factors raised to the same power

$\left(u^2x^2+2x^2u\right)dx-ux^2\cdot dx-x^{3}du=0$

Multiply the single term $dx$ by each term of the polynomial $\left(u^2x^2+2x^2u\right)$

$u^2x^2dx+2x^2u\cdot dx-ux^2\cdot dx-x^{3}du=0$

Combining like terms $2x^2u\cdot dx$ and $-ux^2\cdot dx$

$u^2x^2dx+u\cdot x^2\cdot dx-x^{3}du=0$

Group the terms of the equation

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

Multiply both sides of the equation by $-1$

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

Factor the polynomial $u^2x^2dx+u\cdot x^2\cdot dx$ by it's greatest common factor (GCF): $u\cdot x^2\cdot dx$

$x^{3}du=u\cdot x^2\left(u+1\right)\cdot dx$

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

$\frac{1}{u}\frac{1}{u+1}du=\frac{x^2}{x^{3}}dx$

Simplify the expression $\frac{1}{u}\frac{1}{u+1}du$

$\frac{1}{u\left(u+1\right)}du=\frac{x^2}{x^{3}}dx$

Simplify the expression $\frac{x^2}{x^{3}}dx$

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

Expand and simplify

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

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

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

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

$\frac{1}{u}+\frac{-1}{u+1}$

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

$\int\frac{1}{u}du+\int\frac{-1}{u+1}du$

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

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

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 $n=-1$

$\ln\left|u\right|-\ln\left|u+1\right|$
6

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

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

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

$\ln\left|u\right|-\ln\left|u+1\right|=\ln\left|x\right|+C_0$
8

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

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

Réponse finale au problème

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

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