dont know how to start on thee 2 differential equations

**zen-x** · May 26th 2009, 10:17 AM

Please help, I need to solve this differential equation:

dy/dx + xy = x^3 , f(0)=1

AND

(e^(xy))(1+xy)dx + (x^2)(e^(xy))dy = 0

I cannot seem to separate the variables in the first one and can't find a homogeneous solution and the second one I just had a mental blank...
please show me how to solve step by step?
thanks

**pickslides** · May 26th 2009, 01:59 PM

Originally Posted by zen-x

dy/dx + xy = x^3 , f(0)=1

You need to use the integrating factor method on this equation as it is not separable.

for an equation in the from $y'+f(x)y = g(x)$

your integrating factor is found as follows

$I = e^{\int f(x) dx} = e^{\int x dx} =e^{ \frac{x^2}{2}}$

multiplying this through the entire equation gives

$e^{ \frac{x^2}{2}}y' + xe^{ \frac{x^2}{2}}y = x^3 e^{ \frac{x^2}{2}}$

then using the product rule, LHS changes to be

$(e^{ \frac{x^2}{2}}y)' = x^3 e^{ \frac{x^2}{2}}$

now integrate both sides

$e^{ \frac{x^2}{2}}y = \int x^3 e^{ \frac{x^2}{2}}dx$

gives y to be

$y = \frac{\int x^3 e^{ \frac{x^2}{2}}dx}{e^{ \frac{x^2}{2}}}$

you will have to finish the problem by using integration by parts on the numerator of the RHS.

use $f(0) = 1$ to determine your constant of integration afterwards.

**HallsofIvy** · May 26th 2009, 02:24 PM

Originally Posted by pickslides

You need to use the integrating factor method on this equation as it is not separable.

for an equation in the from $y'+f(x)y = g(x)$

your integrating factor is found as follows

$I = e^{\int f(x) dx} = e^{\int x dx} =e^{ \frac{x^2}{2}}$

multiplying this through the entire equation gives

$e^{ \frac{x^2}{2}}y' + xe^{ \frac{x^2}{2}}y = x^3 e^{ \frac{x^2}{2}}$

then using the product rule, LHS changes to be

$(e^{ \frac{x^2}{2}}y)' = x^3 e^{ \frac{x^2}{2}}$

now integrate both sides

$e^{ \frac{x^2}{2}}y = \int x^3 e^{ \frac{x^2}{2}}dx$

gives y to be

$y = \frac{\int x^3 e^{ \frac{x^2}{2}}dx}{e^{ \frac{x^2}{2}}}$

you will have to finish the problem by using integration by parts on the numerator of the RHS.

First write that as $\int (x^2e^{\frac{x^2}{2}})(x dx)$ and let $u= x^2$ . Then use integration by parts.

use $f(0) = 1$ to determine your constant of integration afterwards.

**pickslides** · May 26th 2009, 03:19 PM

Originally Posted by zen-x

(e^(xy))(1+xy)dx + (x^2)(e^(xy))dy = 0

you need to do a little algebra on this one to put it in the correct form.

$e^{xy}(1+xy)dx + x^2e^{xy}dy = 0$

$e^{xy}(1+xy)dx = -x^2e^{xy}dy$

$e^{xy}(1+xy)dx = -x^2e^{xy}dy$

divide both sides by $e^{xy}$

$(1+xy)dx = -x^2dy$

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

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

$\frac{dy}{dx} = \frac{1}{-x^2}+\frac{xy}{-x^2}$

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

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

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

now you can use the integrating factor method again as above

where $I = e^{\int\frac{1}{x}dx}$

good luck!

**zen-x** · June 11th 2009, 07:55 AM

Originally Posted by pickslides

You need to use the integrating factor method on this equation as it is not separable.

for an equation in the from $y'+f(x)y = g(x)$

your integrating factor is found as follows

$I = e^{\int f(x) dx} = e^{\int x dx} =e^{ \frac{x^2}{2}}$

multiplying this through the entire equation gives

$e^{ \frac{x^2}{2}}y' + xe^{ \frac{x^2}{2}}y = x^3 e^{ \frac{x^2}{2}}$

then using the product rule, LHS changes to be

$(e^{ \frac{x^2}{2}}y)' = x^3 e^{ \frac{x^2}{2}}$

now integrate both sides

$e^{ \frac{x^2}{2}}y = \int x^3 e^{ \frac{x^2}{2}}dx$

gives y to be

$y = \frac{\int x^3 e^{ \frac{x^2}{2}}dx}{e^{ \frac{x^2}{2}}}$

you will have to finish the problem by using integration by parts on the numerator of the RHS.

use $f(0) = 1$ to determine your constant of integration afterwards.

Thanks so much for the help, however, just one thing that i could not understand in the algebra here:

then using the product rule, LHS changes to be

$(e^{ \frac{x^2}{2}}y)' = x^3 e^{ \frac{x^2}{2}}$

How the the LHS transform like that?

**HallsofIvy** · June 11th 2009, 11:45 AM

In a sense, he is using the product rule "in reverse". The product rule states that (uv)'= u'v+ uv'.

Here, you have $e^{ \frac{x^2}{2}}y' + xe^{ \frac{x^2}{2}}y$
Further, it is easy to see that $\left(e^{\frac{x^2}{2}}\right)'= xe^{\frac{x^2}{2}}$ so take $u= e^{\frac{x^2}{2}}$ and v= y.

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