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# Separable Differential Equation Calculator

## Get detailed solutions to your math problems with our Separable Differential Equation step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here.

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###  Difficult Problems

1

Solved example of separable differential equation

$\frac{dy}{dx}=y^2-4$
2

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

$\frac{1}{y^2-4}dy=dx$
3

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

$\int\frac{1}{y^2-4}dy=\int1dx$

Rewrite the expression $\frac{1}{y^2-4}$ inside the integral in factored form

$\int\frac{1}{\left(y+2\right)\left(y-2\right)}dy$

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

$\frac{1}{\left(y+2\right)\left(y-2\right)}=\frac{A}{y+2}+\frac{B}{y-2}$

Find the values for the unknown coefficients: $A, B$. The first step is to multiply both sides of the equation from the previous step by $\left(y+2\right)\left(y-2\right)$

$1=\left(y+2\right)\left(y-2\right)\left(\frac{A}{y+2}+\frac{B}{y-2}\right)$

Multiplying polynomials

$1=\frac{\left(y+2\right)\left(y-2\right)A}{y+2}+\frac{\left(y+2\right)\left(y-2\right)B}{y-2}$

Simplifying

$1=\left(y-2\right)A+\left(y+2\right)B$

Expand the polynomial

$1=\left(y-2\right)A+\left(y+2\right)B$

Assigning values to $y$ we obtain the following system of equations

$\begin{matrix}1=-4A&\:\:\:\:\:\:\:(y=-2) \\ 1=4B&\:\:\:\:\:\:\:(y=2)\end{matrix}$

Proceed to solve the system of linear equations

$\begin{matrix} -4A & + & 0B & =1 \\ 0A & + & 4B & =1\end{matrix}$

Rewrite as a coefficient matrix

$\left(\begin{matrix}-4 & 0 & 1 \\ 0 & 4 & 1\end{matrix}\right)$

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & -\frac{1}{4} \\ 0 & 1 & \frac{1}{4}\end{matrix}\right)$

The integral of $\frac{1}{\left(y+2\right)\left(y-2\right)}$ in decomposed fraction equals

$\int\left(\frac{-1}{4\left(y+2\right)}+\frac{1}{4\left(y-2\right)}\right)dy$

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

$\int\frac{-1}{4\left(y+2\right)}dy+\int\frac{1}{4\left(y-2\right)}dy$

Take the constant $\frac{1}{4}$ out of the integral

$\frac{1}{4}\int\frac{-1}{y+2}dy+\int\frac{1}{4\left(y-2\right)}dy$

Take the constant $\frac{1}{4}$ out of the integral

$\frac{1}{4}\int\frac{-1}{y+2}dy+\frac{1}{4}\int\frac{1}{y-2}dy$

Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left(x+b\right)+C, where b=2, x=y and n=-1 -\frac{1}{4}\ln\left(y+2\right)+\frac{1}{4}\int\frac{1}{y-2}dy Apply the formula: \int\frac{n}{x+b}dx$$=n\ln\left(x+b\right)+C$, where $b=-2$, $x=y$ and $n=1$

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

Solve the integral $\int\frac{1}{y^2-4}dy$ and replace the result in the differential equation

$-\frac{1}{4}\ln\left(y+2\right)+\frac{1}{4}\ln\left(y-2\right)=\int1dx$

The integral of a constant is equal to the constant times the integral's variable

$x$

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

$x+C_0$
5

Solve the integral $\int1dx$ and replace the result in the differential equation

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

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