# Thread: This Bernouilli ODE is killing me

1. ## This Bernouilli ODE is killing me

I've been working on this one differential equation for an hour and a half now and I can't for the life of me figure out where I'm going wrong: $\displaystyle xy' -15 y = (9 x^2 -5 x + 2) y^{\frac{4}{5}},\qquad y(1)=0$

First I divide by $\displaystyle x$. Then by $\displaystyle y^{\frac{4}{5}}$.

This leaves me with:
f(x) = (9x^ - 5x + 2)

y^(-4/5)y' - (15/x)y^(1/5) = f(x)/x

I substitute:
v = y^(1/5)
v' = (y^(-4/5)y')/5

==> 5v' - (15/x)*v = f(x)/x
==> v' - (3/x)*v = f(x)/(5x) [Linear Eqn]

Find integrating factor:
mu = e^(I(-3/x)) = x^-3

So I have:
(v*x^-3)' = 9/(5x^2) - (x^-3) + 2/(5x^4)

Integrate:
v*x^-3 = -9/5x + 1/(2x^2) - 2/(15x^3) + c
v = (-9x^2)/5 + x/2 - 2/15 + Cx^3 ===> g(x)
y^(1/5) = g(x)
y = g(x) ^ 5

Now to solve the IVP I simply y = 0 and x = 1[

Leaves me with:
0 = -9/5 + 1/2 -2/15 + C
C = -43/30

y = ((-9x^2)/5 + x/2 - 2/15 - (43x^3/30))^5 is my final solution

Which when I plug it into the answer box I am told is incorrect. I feel like I must be missing something horribly obvious, and will probably injure my forehead once someone explains to me where I'm going wrong.

2. Originally Posted by BigC
I've been working on this one differential equation for an hour and a half now and I can't for the life of me figure out where I'm going wrong: $\displaystyle xy' -15 y = (9 x^2 -5 x + 2) y^{\frac{4}{5}},\qquad y(1)=0$

First I divide by $\displaystyle x$. Then by $\displaystyle y^{\frac{4}{5}}$.

This leaves me with:
f(x) = (9x^ - 5x + 2)

y^(-4/5)y' - (15/x)y^(1/5) = f(x)/x

I substitute:
v = y^(1/5)
v' = (y^(-4/5)y')/5

==> 5v' - (15/x)*v = f(x)/x
==> v' - (3/x)*v = f(x)/(5x) [Linear Eqn]

Find integrating factor:
mu = e^(I(-3/x)) = x^-3

So I have:
(v*x^-3)' = 9/(5x^2) - (x^-3) + 2/(5x^4)

Integrate:
v*x^-3 = -9/5x + 1/(2x^2) - 2/(15x^3) + c
v = (-9x^2)/5 + x/2 - 2/15 + Cx^3 ===> g(x)
y^(1/5) = g(x)
y = g(x) ^ 5

Now to solve the IVP I simply y = 0 and x = 1[

Leaves me with:
0 = -9/5 + 1/2 -2/15 + C
C = -43/30

y = ((-9x^2)/5 + x/2 - 2/15 - (43x^3/30))^5 is my final solution

Which when I plug it into the answer box I am told is incorrect. I feel like I must be missing something horribly obvious, and will probably injure my forehead once someone explains to me where I'm going wrong.
Divide through by x to get $\displaystyle \frac{\,dy}{\,dx}-\frac{15}{x}y=\left(9x-5+\frac{2}{x}\right)y^{\frac{4}{5}}$

Now apply the substitution $\displaystyle z=y^{1-\frac{4}{5}}=y^{\frac{1}{5}}\implies z^{5}=y$

Thus, $\displaystyle \frac{\,dy}{\,dx}=\frac{\,dy}{\,dz}\cdot\frac{\,dz }{\,dx}=5z^{4}\frac{\,dz}{\,dx}$

Thus, the DE becomes:

$\displaystyle 5z^{4}\frac{\,dz}{\,dx}-\frac{15}{x}z^{5}=\left(9x-5+\frac{2}{x}\right)z^{4}\implies\frac{\,dz}{\,dx}-\frac{3}{x}z=\left(\tfrac{9}{5}x-1+\tfrac{2}{5}x^{-1}\right)$

Now we see its linear.

The integrating factor is $\displaystyle \varrho\!\left(x\right)=x^{-3}$

Thus the DE becomes $\displaystyle \frac{\,d}{\,dx}\left[x^{-3}z\right]=\left(\tfrac{9}{5}x^{-2}-x^{-3}+\tfrac{2}{5}x^{-4}\right)$

Thus, we see that $\displaystyle x^{-3}z=-\tfrac{9}{5}x^{-1}+\tfrac{1}{2}x^{-2}-\tfrac{2}{15}x^{-3}+C\implies z=-\tfrac{9}{5}x^2+\tfrac{1}{2}x-\tfrac{2}{15}+C$

Now apply the initial condition to find that $\displaystyle 0=-\tfrac{9}{5}+\tfrac{1}{2}-\tfrac{2}{15}+C\implies C=\tfrac{43}{30}$

Thus, $\displaystyle z=-\tfrac{9}{5}x^2+\tfrac{1}{2}x-\tfrac{2}{15}+\tfrac{43}{30}$

Now, since $\displaystyle z=y^{\frac{1}{5}}$ we see that the Solution is $\displaystyle y^{\frac{1}{5}}=-\tfrac{9}{5}x^2+\tfrac{1}{2}x-\tfrac{2}{15}+\tfrac{43}{30}$ $\displaystyle \implies \color{red}\boxed{y=\left(-\tfrac{9}{5}x^2+\tfrac{1}{2}x-\tfrac{2}{15}+\tfrac{43}{30}\right)^5}$

Does this make sense?

3. Your solution is virtually identical to mine.

I don't understand how the final solution is ^3 and not to ^5

4. Originally Posted by BigC
Your solution is virtually identical to mine.

I don't understand how the final solution is ^3 and not to ^5
Thanks for catching that. I did mean ^5, not ^3

Note that there is a difference in our solutions!!! I don't have $\displaystyle \frac{43}{30}$ multiplied to $\displaystyle x^{3}$!!

5. Originally Posted by BigC
Integrate:
v*x^-3 = -9/5x + 1/(2x^2) - 2/(15x^3) + c
v = (-9x^2)/5 + x/2 - 2/15 + Cx^3 ===> g(x)
y^(1/5) = g(x)
y = g(x) ^ 5

Now to solve the IVP I simply y = 0 and x = 1[

Leaves me with:
0 = -9/5 + 1/2 -2/15 + C
C = -43/30
I think your problem is here. c = +43/30

Originally Posted by BigC
y = ((-9x^2)/5 + x/2 - 2/15 - (43x^3/30))^5 is my final solution

Which when I plug it into the answer box I am told is incorrect. I feel like I must be missing something horribly obvious, and will probably injure my forehead once someone explains to me where I'm going wrong.
Try

$\displaystyle y = \left(\frac{43}{30}x^3 - \frac{9}{5}x^2 + \frac{1}{2}x - \frac{2}{15}\right)^5$