# Solving differential equation

• Mar 25th 2009, 01:49 AM
LooNiE
Solving differential equation
The equation I have is:

(x+1)dy/dx - y = x^2 - 1

I basically need to rearrange it to find the integrating factor, then I need to solve it to find the general solution. Also how do I solve the particular solution? I need to find it when y(0) = 0.

Thanks for any help

LooN
• Mar 25th 2009, 02:49 AM
Craka
The equation i read as being

$
(x + 1)\frac{{dy}}{{dx}} - y = x^2 - 1
$

Linear differential equation needs to be in the form $y' + P(x)y = Q(x)$

So divide through by x+1 to get dy/dx on its own.

$
\begin{array}{l}
\frac{{dy}}{{dx}} - \left( {\frac{1}{{x + 1}}} \right)y = \frac{{x^2 - 1}}{{x + 1}} \\
\frac{{dy}}{{dx}} - \left( {\frac{1}{{x + 1}}} \right)y = \frac{{(x - 1)(x + 1)}}{{x + 1}} \\
\frac{{dy}}{{dx}} - \left( {\frac{1}{{x + 1}}} \right)y = x - 1 \\
\end{array}
$

The integrating factor is in the form $I(x) = e^{\int {P(x)dx} }$

$
\begin{array}{l}
P(x) = \frac{1}{{x + 1}} \\
Q(x) = x - 1 \\
\end{array}
$

So integrating factor will be
$
I(x) = e^{\int {P(x)dx} } = e^{\int {\frac{1}{{x + 1}}dx} } = e^{\ln (x + 1)}
$

From this you need to multiply the differential equation in linear form by I(x)
ie $I(x)y' + I(x)P(x)y = I(x)Q(x)$

From here I get a little lost myself, I'm learning these also, so I'll leave it here so that maybe someone can clarify the rest.
• Mar 25th 2009, 06:08 AM
LooNiE
further help
can anyone help out further with this question? Its so confusing. !bump
• Mar 25th 2009, 08:01 AM
LooNiE
!bump
• Mar 25th 2009, 08:25 AM
chisigma
It is a linear differential equation that has 'standard form' ...

$y'= a(x)\cdot y + b(x)$

... and its solution is...

$y= e^{\int a(x)\cdot dx}\cdot [\int b(x)\cdot e^{- \int a(x)\cdot dx} + c]$

In our case is...

$a(x)=\frac{1}{1+x}$

$b(x)=x - 1$

... so that...

$\int a(x)\cdot dx = \int \frac{dx}{1+x}= \ln (1+x)$

$\int b(x)\cdot e^{- \int a(x)\cdot dx}= - \int \frac{1-x}{1+x}\cdot dx= x -2\cdot \ln (1+x)$

... and the complete solution is...

$y= (1+x)\cdot \{x -2\cdot \ln (1+x) + c \}$

Kind regards

$\chi$ $\sigma$
• Mar 25th 2009, 02:09 PM
tom@ballooncalculus
Just in case a picture helps...

Once we have the equation in this form:

$\frac{dy}{dx}\ +\ y\ \frac{-1}{x+1}\ =\ x-1$

we then want to make it fit somehow into the balloons along the bottom row here...

http://www.ballooncalculus.org/asy/i...erXPlusOne.png

... where the triangular network satisfies the product rule in this pattern...

http://www.ballooncalculus.org/prod_rule.png

So far our attempt fails because we'd need to carry on by entering the integral of $\frac{-1}{x+1}$ in each of the two remaining spaces in the product-rule shape.

Which would be fine if we could work out how to modify x-1 on the right, accordingly, but that doesn't look easy.

The trick is to treat $\frac{-1}{x+1}$ as the derivative 'by-product' of a chain-rule differentiation.

The straight dashed line here integrates/differentiates (up/down) with respect to whatever expression we put in the dashed balloon -

i.e. the 'inner' function, which will be the integral of $\frac{-1}{x+1}$ ...

http://www.ballooncalculus.org/asy/i...rXPlusOne1.png

... giving us

$y\ \frac{1}{x+1}\ =\ 2\ -\ 2\ \ln|x+1|\ +\ c$

Taking the exponential of the inner function does the trick because we can multiply it by each term in the equation, and the product-rule is satisfied.

We've removed some of the balloons of the product-rule shape, but this is a matter of taste, and sometimes space...

http://www.ballooncalculus.org/asy/i...rXPlusOne2.png

Don't integrate - balloontegrate!

Balloon Calculus: worked examples from past papers