# Substituting

• Oct 5th 2009, 07:34 PM
I-Think
Substituting
Show that, with a suitable value of the constant $\displaystyle \alpha$, the substitution $\displaystyle y=x^{\alpha}w$ reduces the differential equation

$\displaystyle 2x^{2}\frac{d^{2}y}{dx^2}+(3x^2+8x)\frac{dy}{dx}+( x^2+6x+4)=f(x)$

to

$\displaystyle 2\frac{d^{2}w}{dx^2}+3\frac{dw}{dx}+w=f(x)$

Preparing for an exam,so all help is greatly appreciated

Edit
Ooops.Forgot the y
The expression is
$\displaystyle 2x^{2}\frac{d^{2}y}{dx^2}+(3x^2+8x)\frac{dy}{dx}+( x^2+6x+4)y=f(x)$
• Oct 5th 2009, 11:44 PM
mr fantastic
Quote:

Originally Posted by I-Think
Show that, with a suitable value of the constant $\displaystyle \alpha$, the substitution $\displaystyle y=x^{\alpha}w$ reduces the differential equation

$\displaystyle 2x^{2}\frac{d^{2}y}{dx^2}+(3x^2+8x)\frac{dy}{dx}+( x^2+6x+4)=f(x)$

to

$\displaystyle 2\frac{d^{2}w}{dx^2}+3\frac{dw}{dx}+w=f(x)$

Preparing for an exam,so all help is greatly appreciated

Get $\displaystyle \frac{dy}{dx}$ using the product rule. Get $\displaystyle \frac{d^2y}{dx^2}$ by using the product rule to differentiate your expression for $\displaystyle \frac{dy}{dx}$.

Substitute your expressions for $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d^2y}{dx^2}$ into the DE and simplify.

If you need more help please show what you've done and say where you get stuck.
• Oct 6th 2009, 06:00 AM
I-Think
Still in a spot of bother over the differentiation process.

If $\displaystyle y=x^{\alpha}w$
Using the product rule

$\displaystyle \frac{dy}{dx}=\alpha{wx^{\alpha-1}}+x^{\alpha}{\frac{dw}{dx}}$

And

$\displaystyle \frac{d^{2}y}{dx^2}=\alpha{x^{\alpha-1}}\frac{dw}{dx}+w(\alpha^2-\alpha)x^{\alpha-2}+\alpha{x^{\alpha-1}}\frac{dw}{dx}+x^\alpha{\frac{d^2w}{dx^2}}$

When I substitute this into the equation, I'm not getting the correct answer. May I have more guidance please?
• Oct 6th 2009, 02:29 PM
mr fantastic
Quote:

Originally Posted by I-Think
Still in a spot of bother over the differentiation process.

If $\displaystyle y=x^{\alpha}w$
Using the product rule

$\displaystyle \frac{dy}{dx}=\alpha{wx^{\alpha-1}}+x^{\alpha}{\frac{dw}{dx}}$

And

$\displaystyle \frac{d^{2}y}{dx^2}=\alpha{x^{\alpha-1}}\frac{dw}{dx}+w(\alpha^2-\alpha)x^{\alpha-2}+\alpha{x^{\alpha-1}}\frac{dw}{dx}+x^\alpha{\frac{d^2w}{dx^2}}$

When I substitute this into the equation, I'm not getting the correct answer. May I have more guidance please?

Have you checked for careless mistakes?