Here, we show you a step-by-step solved example of homogeneous differential equation. This solution was automatically generated by our smart calculator:
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
Use the substitution: $y=ux$
Factor the polynomial $2x+ux$ by it's greatest common factor (GCF): $x$
Expand the fraction $\frac{u\cdot dx+x\cdot du}{dx}$ into $2$ simpler fractions with common denominator $dx$
Simplify the resulting fractions
Multiply the single term $-1$ by each term of the polynomial $\left(4x+3ux\right)$
Factor the polynomial $-4x-3ux$ by it's greatest common factor (GCF): $-x$
Simplify the fraction $\frac{-x\left(4+3u\right)}{x\left(2+u\right)}$ by $x$
Multiply the single term $-1$ by each term of the polynomial $\left(4+3u\right)$
We need to isolate the dependent variable $u$, we can do that by simultaneously subtracting $u$ from both sides of the equation
Combine all terms into a single fraction with $2+u$ as common denominator
Combining like terms $-3u$ and $-2u$
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
Simplify the expression $\frac{2+u}{-4-5u-u^2}du$
Expand and simplify
Integrate both sides of the differential equation, the left side with respect to $u$, and the right side with respect to $x$
Take the constant $\frac{1}{-1}$ out of the integral
Rewrite the fraction $\frac{2+u}{\left(u+1\right)\left(u+4\right)}$ in $2$ simpler fractions using partial fraction decomposition
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
Take the constant $\frac{1}{3}$ out of the integral
Multiply the fraction and term in $- \left(\frac{1}{3}\right)\int\frac{1}{u+1}du$
Take the constant $\frac{1}{3}$ out of the integral
Multiply the fraction and term in $- \left(\frac{1}{3}\right)\int\frac{2}{u+4}du$
Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=1$, $x=u$ and $n=1$
Any expression multiplied by $1$ is equal to itself
Apply the formula: $\int\frac{n}{x+b}dx$$=nsign\left(x\right)\ln\left(x+b\right)+C$, where $b=4$, $x=u$ and $n=2$
Multiply the fraction and term in $2\left(-\frac{1}{3}\right)\ln\left|u+4\right|$
Solve the integral $\int\frac{2+u}{-\left(u+1\right)\left(u+4\right)}du$ and replace the result in the differential equation
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
Solve the integral $\int\frac{1}{x}dx$ and replace the result in the differential equation
Replace $u$ with the value $\frac{y}{x}$
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