# Separable 1st order ODE

• May 7th 2010, 06:36 AM
aspotonthewall
Separable 1st order ODE
I've already solved this problem, but the assessment system I'm supposed to insert my answer to keeps telling me it's incorrect, so I must have made a mistake somewhere and can't spot it.

So I'll write out, step by step, what I've done:

$y''-21-7x^2+3y^2+x^2y^2=0$

$=> y''=(7-y^2)(x^2+3)$

$=>\int{\frac{1}{7-y^2} dy}=\int{x^2+3 dx}$

Right Hand Side:

$\int{x^2+3 dx}=\frac{x^3}{3}+3x+c$

Left Hand Side:

Rewriting LHS using partial fractions:

$\frac{1}{7-y^2}= \frac {A}{\sqrt{7}-y} + \frac{B}{\sqrt{7}+y}=\frac{A(\sqrt{7}+y)}{(\sqrt{7 }-y)(\sqrt{7}+y)}+\frac{B(\sqrt{7}-y)}{(\sqrt{7}-y)(\sqrt{7}+y)}$ $=\frac{A(\sqrt{7}+y)+B(\sqrt{7}-y)}{7-y^2}$

$A(\sqrt{7}+y)+B(\sqrt{7}-y)=1$

Let $y=\sqrt{7}$
$A(\sqrt{7}+\sqrt{7})+B(\sqrt{7}-\sqrt{7})=1$
$=> A(2\sqrt{7})=1$
$=> A=\frac{1}{2\sqrt{7}}$

Let $y=-\sqrt{7}$
$A(\sqrt{7}-\sqrt{7})+B(\sqrt{7}+\sqrt{7})=1$
$=>B(2\sqrt{7})=1$
$=> B=\frac{1}{2\sqrt{7}}$

So then I get LHS to be:
$\frac{1}{2\sqrt{7}}(\frac {1}{\sqrt{7}-y} + \frac{1}{\sqrt{7}+y})$

Integrating the LHS with respect to y gives
$\frac{1}{14}\sqrt{7}\ln{(\sqrt{7}+y)}-\frac{1}{14}\sqrt{7}\ln{(\sqrt{7}-y)}+c$

So now I have
$\frac{1}{14}\sqrt{7}\ln{(\frac{(\sqrt{7}+y)}{\sqrt {7}-y)})}=\frac{x^3}{3}+3x$

$=>y=\sqrt{7}\frac{1-\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{1+\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}+c$

But apparently that's the wrong answer and I can't, for the life of me, figure out where I'm going wrong.
• May 7th 2010, 07:16 AM
Jester
Quote:

Originally Posted by aspotonthewall
I've already solved this problem, but the assessment system I'm supposed to insert my answer to keeps telling me it's incorrect, so I must have made a mistake somewhere and can't spot it.

So I'll write out, step by step, what I've done:

$y''-21-7x^2+3y^2+x^2y^2=0$

$=> y''=(7-y^2)(x^2+3)$

$=>\int{\frac{1}{7-y^2} dy}=\int{x^2+3 dx}$

Right Hand Side:

$\int{x^2+3 dx}=\frac{x^3}{3}+3x+c$

Left Hand Side:

Rewriting LHS using partial fractions:

$\frac{1}{7-y^2}= \frac {A}{\sqrt{7}-y} + \frac{B}{\sqrt{7}+y}=\frac{A(\sqrt{7}+y)}{(\sqrt{7 }-y)(\sqrt{7}+y)}+\frac{B(\sqrt{7}-y)}{(\sqrt{7}-y)(\sqrt{7}+y)}$ $=\frac{A(\sqrt{7}+y)+B(\sqrt{7}-y)}{7-y^2}$

$A(\sqrt{7}+y)+B(\sqrt{7}-y)=1$

Let $y=\sqrt{7}$
$A(\sqrt{7}+\sqrt{7})+B(\sqrt{7}-\sqrt{7})=1$
$=> A(2\sqrt{7})=1$
$=> A=\frac{1}{2\sqrt{7}}$

Let $y=-\sqrt{7}$
$A(\sqrt{7}-\sqrt{7})+B(\sqrt{7}+\sqrt{7})=1$
$=>B(2\sqrt{7})=1$
$=> B=\frac{1}{2\sqrt{7}}$

So then I get LHS to be:
$\frac{1}{2\sqrt{7}}(\frac {1}{\sqrt{7}-y} + \frac{1}{\sqrt{7}+y})$

Integrating the LHS with respect to y gives
$\frac{1}{14}\sqrt{7}\ln{(\sqrt{7}+y)}-\frac{1}{14}\sqrt{7}\ln{(\sqrt{7}-y)}+c$

So now I have
$\frac{1}{14}\sqrt{7}\ln{(\frac{(\sqrt{7}+y)}{\sqrt {7}-y)})}=\frac{x^3}{3}+3x$

$=>y=\sqrt{7}\frac{1-\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{1+\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}+c$

But apparently that's the wrong answer and I can't, for the life of me, figure out where I'm going wrong.

You have $+ c$. I think this is wrong. From what you have

$\frac{1}{14}\sqrt{7}\ln{(\frac{(\sqrt{7}+y)}{\sqrt {7}-y)})}=\frac{x^3}{3}+3x + \frac{1}{14}\sqrt{7}\, \ln c$ (the last part is the arbitrary constant). Solving for $y$ gives

$y=\sqrt{7}\,\frac{c-\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{c+\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}$ (Note where my c's are.)
• May 7th 2010, 07:26 AM
aspotonthewall
I just tried it and entered it into the assessment system, and it's still wrong, but the error I got was:

"Your answer does satisfy the differential equation but either does not have a constant specified or you did not use "c" (small case) as instructed. Your mark for this attempt is 0."
• May 7th 2010, 07:37 AM
Jester
Quote:

Originally Posted by aspotonthewall
I just tried it and entered it into the assessment system, and it's still wrong, but the error I got was:

"Your answer does satisfy the differential equation but either does not have a constant specified or you did not use "c" (small case) as instructed. Your mark for this attempt is 0."

$
y=\sqrt{7}\,\frac{1-c\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{1+c\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}$
?
• May 7th 2010, 07:58 AM
aspotonthewall
I get the exact same error :(

Oh well... thanks anyway!
• May 7th 2010, 08:25 AM
Jester
Quote:

Originally Posted by aspotonthewall
I get the exact same error :(

Oh well... thanks anyway!

One last attempt. Try

$
y=\sqrt{7}\,\frac{\exp{(\frac{2}{3}\sqrt{7}x^3+6\s qrt{7}x + c)} - 1}{\exp{(\frac{2}{3}\sqrt{7}x^3+6\sqrt{7}x+c)}+1}$
.

Also try

$
y = \sqrt{7} \tanh \left( 3 \sqrt{7} x + \frac{\sqrt{7}}{3} x^3 + \sqrt{7} c_1 \right)
$

This is what Maple gave as the solution.