Question

differential equation using separable variable method of dy/y=dx/x^2(1-x^2)=0

Answer 1

Step by step solutions
http://www.mathskey.com/question2answer/4234/differential-equation-using-separable-variable-method-of

Answer 2

\[\frac{dy}{y}=\frac{dx}{x^2(1-x^2)}\]
Integrate both sides:
\[\int\frac{dy}{y}=\int\frac{dx}{x^2(1-x^2)}\]
Split the right hand side into partial fractions:
\[\frac{1}{x^2(1-x^2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{1-x}+\frac{D}{1+x}\\
1=Ax(1-x^2)+B(1-x^2)+Cx^2(1+x)+Dx^2(1-x)\\
1=Ax-Ax^3+B-Bx^2+Cx^2+Cx^3+Dx^2-Dx^3\\
1=(-A+C-D)x^3+(-B+C+D)x^2+Ax+B\]
yielding the system
\[\begin{cases}
-A+C-D=0\\-B+C+D=0\\A=0\\B=1
\end{cases}~\Rightarrow~A=0,B=1,C=\frac{1}{2},D=\frac{1}{2}\]
So your equation becomes
\[\int\frac{dy}{y}=\int\left(\frac{1}{x^2}+\frac{1}{2(1-x)}+\frac{1}{2(1+x)}\right)~dx\\
\int\frac{dy}{y}=\int\frac{dx}{x^2}+\frac{1}{2}\int\frac{dx}{1-x}+\frac{1}{2}\int\frac{dx}{1+x}\\
\ln|y|=-\frac{1}{x} +\frac{1}{2}\ln|1-x|+\frac{1}{2}\ln|1+x|+C
\]