# inhomogeneous linear differential equation

• Feb 1st 2009, 10:38 AM
rajr
inhomogeneous linear differential equation
b)
Determine the general solution of the inhomogeneous linear differential equation

y′ = [xy / (1+x^2)] + SQRT [ (1+x^2) / (1-x^2)]
by the method of integrating factor.
• Feb 1st 2009, 11:22 AM
Chris L T521
Quote:

Originally Posted by rajr
b)
Determine the general solution of the inhomogeneous linear differential equation

y′ = [xy / (1+x^2)] + SQRT [ (1+x^2) / (1-x^2)]
by the method of integrating factor.

First, rewrite the DE as follows: $\frac{\,dy}{\,dx}-\frac{x}{1+x^2}y=\sqrt{\frac{1+x^2}{1-x^2}}$

The integrating factor would be $\varrho\!\left(x\right)=e^{-\int\frac{x}{1+x^2}\,dx}$

By u-substitution, it can be shown that $\varrho\!\left(x\right)=\frac{1}{\sqrt{1+x^2}}$

Thus, the DE becomes $\frac{\,d}{\,dx}\left[\frac{1}{\sqrt{1+x^2}}\cdot y\right]=\frac{1}{\sqrt{1-x^2}}$

It should be pretty straightforward from here.

Does this make sense?
• Feb 1st 2009, 11:32 AM
rajr
hi thx for ur help

should i then integrate both sides to get Y
• Feb 1st 2009, 11:37 AM
anon18
no you just integrate the left side and you should be left with an answer like y=(1+x^2)^-1/2[sin^-1(x)+C]
• Feb 1st 2009, 11:50 AM
rajr
thx for ur help
• Feb 1st 2009, 11:54 AM
Chris L T521
Quote:

Originally Posted by anon18
no you just integrate the left side and you should be left with an answer like y=(1+x^2)^-1/2[sin^-1(x)+C]

No. You integrate both sides to get $\frac{1}{\sqrt{1+x^2}}\cdot y=\sin^{-1}\!\left(x\right)+C\implies y=\sqrt{1+x^2}\left[\sin^{-1}\!\left(x\right)+C\right]$ $\implies y=\sqrt{1+x^2}\sin^{-1}\!\left(x\right)+C\sqrt{1+x^2}$.

When writing the solution, though, we usually prefer to write it as $y=\sqrt{1+x^2}\left[\sin^{-1}\!\left(x\right)+C\right]$
• Feb 1st 2009, 12:10 PM
rajr
thx for ur help Mr Chris....