# Thread: Separable 1st order ODE

1. ## 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:

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

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

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

Right Hand Side:

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

Left Hand Side:

Rewriting LHS using partial fractions:

$\displaystyle \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)}$$\displaystyle =\frac{A(\sqrt{7}+y)+B(\sqrt{7}-y)}{7-y^2} \displaystyle A(\sqrt{7}+y)+B(\sqrt{7}-y)=1 Let \displaystyle y=\sqrt{7} \displaystyle A(\sqrt{7}+\sqrt{7})+B(\sqrt{7}-\sqrt{7})=1 \displaystyle => A(2\sqrt{7})=1 \displaystyle => A=\frac{1}{2\sqrt{7}} Let \displaystyle y=-\sqrt{7} \displaystyle A(\sqrt{7}-\sqrt{7})+B(\sqrt{7}+\sqrt{7})=1 \displaystyle =>B(2\sqrt{7})=1 \displaystyle => B=\frac{1}{2\sqrt{7}} So then I get LHS to be: \displaystyle \frac{1}{2\sqrt{7}}(\frac {1}{\sqrt{7}-y} + \frac{1}{\sqrt{7}+y}) Integrating the LHS with respect to y gives \displaystyle \frac{1}{14}\sqrt{7}\ln{(\sqrt{7}+y)}-\frac{1}{14}\sqrt{7}\ln{(\sqrt{7}-y)}+c So now I have \displaystyle \frac{1}{14}\sqrt{7}\ln{(\frac{(\sqrt{7}+y)}{\sqrt {7}-y)})}=\frac{x^3}{3}+3x \displaystyle =>y=\sqrt{7}\frac{1-\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{1+\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}+c But apparently that's the wrong answer and I can't, for the life of me, figure out where I'm going wrong. 2. Originally Posted by aspotonthewall I've already solved this problem, but the assessment system I'm supposed to insert my answer to keeps telling me it's incorrect, so I must have made a mistake somewhere and can't spot it. So I'll write out, step by step, what I've done: \displaystyle y''-21-7x^2+3y^2+x^2y^2=0 \displaystyle => y''=(7-y^2)(x^2+3) \displaystyle =>\int{\frac{1}{7-y^2} dy}=\int{x^2+3 dx} Right Hand Side: \displaystyle \int{x^2+3 dx}=\frac{x^3}{3}+3x+c Left Hand Side: Rewriting LHS using partial fractions: \displaystyle \frac{1}{7-y^2}= \frac {A}{\sqrt{7}-y} + \frac{B}{\sqrt{7}+y}=\frac{A(\sqrt{7}+y)}{(\sqrt{7 }-y)(\sqrt{7}+y)}+\frac{B(\sqrt{7}-y)}{(\sqrt{7}-y)(\sqrt{7}+y)}$$\displaystyle =\frac{A(\sqrt{7}+y)+B(\sqrt{7}-y)}{7-y^2}$

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

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

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

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

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

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

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

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

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

$\displaystyle y=\sqrt{7}\,\frac{c-\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{c+\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}$ (Note where my c's are.)

3. 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."

4. 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."

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

5. I get the exact same error

Oh well... thanks anyway!

6. Originally Posted by aspotonthewall
I get the exact same error

Oh well... thanks anyway!
One last attempt. Try

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

Also try

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

This is what Maple gave as the solution.