x^2 y'' - 5x y' - 7y = x^3

**s3a** · November 29th 2011, 09:18 AM

The question asks me to solve x^2 y'' - 5x y' - 7y = x^3 as function of x. In other words, it's asking for y(x). I am given the initial conditions y(1) = 10 and y'(1) = 8.

Someone who wasn't very sure told me to use characteristic equations and the method of undetermined coefficients but is that correct? If so, how do I set it up mechanically? I only know how to use characteristic equations when there are no x variables in the differential equation. I am really, really stuck. I only divided both sides by x^2.

The answer is:
http://www.wolframalpha.com/input/?i=Solve+x^2+y''+-+5xy'+-+7y+%3D+x^3;+y(1)+%3D+10,+y'(1)+%3D+8

If someone could show me how to set up the problem (tedious steps aren't necessary), I would greatly appreciate it!

Thanks in advance!

**TheEmptySet** · November 29th 2011, 09:29 AM

Originally Posted by s3a

The question asks me to solve x^2 y'' - 5x y' - 7y = x^3 as function of x. In other words, it's asking for y(x). I am given the initial conditions y(1) = 10 and y'(1) = 8.

Someone who wasn't very sure told me to use characteristic equations and the method of undetermined coefficients but is that correct? If so, how do I set it up mechanically? I only know how to use characteristic equations when there are no x variables in the differential equation. I am really, really stuck. I only divided both sides by x^2.

The answer is:
http://www.wolframalpha.com/input/?i=Solve+x^2+y''+-+5xy'+-+7y+%3D+x^3;+y(1)+%3D+10,+y'(1)+%3D+8

If someone could show me how to set up the problem (tedious steps aren't necessary), I would greatly appreciate it!

Thanks in advance!

You can use the substitution

$y=f(u)$ where $u=ln(x)$ Then by the chain rule you get

$\frac{dy}{dx}=\frac{df}{du}\frac{du}{dx}=\frac{1}{ x}f'$

Now if you take one more derivative you get

$\frac{d^2y}{dx^2}=\frac{d}{dx}\left( \frac{1}{x}f'\right)=-\frac{1}{x^2}f'+\frac{1}{x}f''\cdot \frac{1}{x}=\frac{1}{x^2}(-f'+f'')$

Now just sub this into the equation and watch the magic happen.

**s3a** · November 29th 2011, 10:58 AM

Sorry for the dumb question but, is this process just a convenient trick or is it how I am supposed to solve most differential equations with x variables in them? Also, is there a name for this process other than just "substitution?"

**s3a** · November 29th 2011, 12:03 PM

Also, I think I almost get it but I get the wrong answer. I'm stuck at evaluating the new initial conditions as well so it would be great if you could help me with that too but I have the answer without the initial conditions plugged in (http://www.wolframalpha.com/input/?i=Solve+x^2+y''+-+5xy'+-+7y+%3D+x^3) and it differs from the answer I get.

I'm attaching my work.

**TheEmptySet** · November 29th 2011, 12:48 PM

The equation is of the type Cauchy- Euler. A Google search will turn up alot of info.

I was a little careless with my notation above and I think it has confused you. When we do this substitution we are chainging the independant variable from x to u. So we need to sub out the x's in the original equation. Using the work above you should get the equation.

$x^2\left[ \frac{1}{x^2}\left( -f'(u)+f''(u)\right)\right]-5x\left( \frac{1}{x}f'(u)\right)-7f(u)=\left( e^{u}\right)^3$

Simplifying will reduce out all of the x's and gives the equation.

Note: here the primes represent differentation with respect to u.

$f''(u)-6'f(u)-7f(u)=e^{3u}$

I hope this clears it up a bit.

TES

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