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.

Printable View

- Feb 1st 2009, 09:38 AMrajrinhomogeneous linear differential equationb)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, 10:22 AMChris L T521
First, rewrite the DE as follows: $\displaystyle \frac{\,dy}{\,dx}-\frac{x}{1+x^2}y=\sqrt{\frac{1+x^2}{1-x^2}}$

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

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

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

It should be pretty straightforward from here.

Does this make sense? - Feb 1st 2009, 10:32 AMrajr
hi thx for ur help

should i then integrate both sides to get Y - Feb 1st 2009, 10:37 AManon18
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, 10:50 AMrajr
thx for ur help

- Feb 1st 2009, 10:54 AMChris L T521
No. You integrate both sides to get $\displaystyle \frac{1}{\sqrt{1+x^2}}\cdot y=\sin^{-1}\!\left(x\right)+C\implies y=\sqrt{1+x^2}\left[\sin^{-1}\!\left(x\right)+C\right]$ $\displaystyle \implies y=\sqrt{1+x^2}\sin^{-1}\!\left(x\right)+C\sqrt{1+x^2}$.

When writing the solution, though, we usually prefer to write it as $\displaystyle y=\sqrt{1+x^2}\left[\sin^{-1}\!\left(x\right)+C\right]$ - Feb 1st 2009, 11:10 AMrajr
thx for ur help Mr Chris....