Reducible Second Order Equation

• February 4th 2009, 04:36 PM
mandy123
Reducible Second Order Equation
Use the Reducible Second Order Equation to find a general solution (Assume x, y, and/or y' positive where helpful)

(x^2)y'' + 3xy'=2

This is what I have but came up with the wrong answer
can someone explain what I need to do (Thanks)

y is missing
Let p= y'= dy/dx then let p'=y''=dp/dx

(x^2) dp/dx +3xp= 2

dp/dx + (3/x)p = (2/x^2)

Now i used the linear first-order equation and found

Dx + (3x^2)p = 2x

I then got

(3x^2)p = x^2 + C

Dividing by (3x^2) I got

p= (1/3) + C/(3x^2)

This is not the right answer, the right answer is ln(x) + Ax^-2 + B
Thanks
• February 4th 2009, 04:48 PM
ThePerfectHacker
Quote:

Originally Posted by mandy123
Use the Reducible Second Order Equation to find a general solution (Assume x, y, and/or y' positive where helpful)

(x^2)y'' + 3xy'=2

You need to remember that you solved for $p$.
The problem asks to solve for $y$. (Surprised)
• February 4th 2009, 04:49 PM
Mush
Quote:

Originally Posted by mandy123
Use the Reducible Second Order Equation to find a general solution (Assume x, y, and/or y' positive where helpful)

(x^2)y'' + 3xy'=2

This is what I have but came up with the wrong answer
can someone explain what I need to do (Thanks)

y is missing
Let p= y'= dy/dx then let p'=y''=dp/dx

(x^2) dp/dx +3xp= 2

dp/dx + (3/x)p = (2/x^2)

Now i used the linear first-order equation and found

Dx + (3x^2)p = 2x

I then got

(3x^2)p = x^2 + C

Dividing by (3x^2) I got

p= (1/3) + C/(3x^2)

This is not the right answer, the right answer is ln(x) + Ax^-2 + B
Thanks

So far you have only found $p$. Remember that $p = \frac{dy}{dx}$. What must you do to $\frac{dy}{dx}$ to get $y$?

But you solved incorrectly for p anyway.

$\frac{dp}{dx} +\frac{3}{x}p = \frac{2}{x^2}$

$\phi = e^{\int \frac{3}{x}dx} = e^{3\ln|x|} = e^{\ln(x^3)} = x^3$

$x^3 p = \int x^3 \times \frac{2}{x^2}dx$

$x^3 p = \int 2xdx$

$x^3 p = x^2+A$

$p = \frac{1}{x}+\frac{A}{x^3}$

$y= \int \frac{1}{x}+\frac{A}{x^3}dx$

$y= \ln|x|-\frac{A}{2x^2}+B$
• February 4th 2009, 05:07 PM
mandy123
http://www.mathhelpforum.com/math-he...36e9ee65-1.gif

Thanks for letting me know I solved for p wrong, I realized my mistake when you said something.
I agree with this answer, but why does the answer key show the answer is ln(x) + Ax^-2 + B
• February 5th 2009, 06:05 AM
Mush
Quote:

Originally Posted by mandy123
http://www.mathhelpforum.com/math-he...36e9ee65-1.gif

Thanks for letting me know I solved for p wrong, I realized my mistake when you said something.
I agree with this answer, but why does the answer key show the answer is ln(x) + Ax^-2 + B

Well taking an arbitrary constant, multiplying it by negative 1 and dividing it by 2 still gives you a an arbitrary constant.

So think of it as:

$y = \ln|x| - \frac{A^*}{2x^2} + B$

$= \ln|x| + Ax^{-2} + B$

Where $A = -\frac{A}{2}$

Yes?
• February 5th 2009, 06:24 AM
mandy123
Ok, sounds good, thank you so much