Homogeneous Differential Equation Calculator

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

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

The integral of a function times a constant ($2$) is equal to the constant times the integral of the function

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

We can solve the integral $\int\frac{1}{u+1}du$ 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 $v$), which when substituted makes the integral easier. We see that $u+1$ it's a good candidate for substitution. Let's define a variable $v$ and assign it to the choosen part

Now, in order to rewrite $du$ in terms of $dv$, we need to find the derivative of $v$. We need to calculate $dv$, we can do that by finding the derivative of the equation above

Substituting $v$ and $du$ in the integral and simplify

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

Replace $v$ with the value that we assigned to it in the beginning: $u+1$

We can solve the integral $\int\frac{1}{4+u}du$ 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 $v$), which when substituted makes the integral easier. We see that $4+u$ it's a good candidate for substitution. Let's define a variable $v$ and assign it to the choosen part

Now, in order to rewrite $du$ in terms of $dv$, we need to find the derivative of $v$. We need to calculate $dv$, we can do that by finding the derivative of the equation above

Substituting $v$ and $du$ in the integral and simplify

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

Replace $v$ with the value that we assigned to it in the beginning: $4+u$

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 $C$