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Homogene Differentialgleichung Rechner

Mit unserem Homogene Differentialgleichung Schritt-für-Schritt-Rechner erhalten Sie detaillierte Lösungen für Ihre mathematischen Probleme. Üben Sie Ihre mathematischen Fähigkeiten und lernen Sie Schritt für Schritt mit unserem Mathe-Löser. Alle unsere Online-Rechner finden Sie hier.

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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:

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

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

Endgültige Antwort auf das Problem

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

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