# Thread: Reducible Second Order Equation

1. ## 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

2. 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 $\displaystyle p$.
The problem asks to solve for $\displaystyle y$.

3. 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 $\displaystyle p$. Remember that $\displaystyle p = \frac{dy}{dx}$. What must you do to $\displaystyle \frac{dy}{dx}$ to get $\displaystyle y$?

But you solved incorrectly for p anyway.

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

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

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

$\displaystyle x^3 p = \int 2xdx$

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

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

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

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

4. 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

5. Originally Posted by mandy123

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:

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

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

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

Yes?

6. Ok, sounds good, thank you so much