MHB Help with another separable equation

[jahrens]

I am really struggling with this one, if anyone can help. (ln(y))3*(dy/dx)=(x^3)y with initial conditions y=e^2 x=1

I get c=4/(e^4) - 1/4
then I get stuck at (3ln^2y - ln^3y)/(y^2)=(x^4)/4 + C Any ideas? I'm really not good at these so there are probably mistakes, because at this point I have no idea how to isolate y.

 

[Prove It]

I am really struggling with this one, if anyone can help. (ln(y))3*(dy/dx)=(x^3)y with initial conditions y=e^2 x=1

I get c=4/(e^4) - 1/4
then I get stuck at (3ln^2y - ln^3y)/(y^2)=(x^4)/4 + C Any ideas? I'm really not good at these so there are probably mistakes, because at this point I have no idea how to isolate y.

Is the DE $\displaystyle \begin{align*} \left[ \ln{(y)} \right] ^3\,\frac{\mathrm{d}y}{\mathrm{d}x} = x^3\,y \end{align*}$?

 

[jahrens]

Yes!

 

[Prove It]

$\displaystyle \begin{align*} \left[ \ln{(y)} \right] ^3\,\frac{\mathrm{d}y}{\mathrm{d}x} &= x^3\,y \\ \left[ \ln{(y)} \right] ^3\,\frac{1}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x} &= x^3 \\ \int{ \left[ \ln{(y)} \right] ^3\,\frac{1}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x} \,\mathrm{d}x} &= \int{ x^3\,\mathrm{d}x} \\ \int{ \left[ \ln{(y)} \right] ^3\,\frac{1}{y}\,\mathrm{d}y} &= \frac{x^4}{4} + C_1 \\ \int{ u^3\,\mathrm{d}u} &= \frac{x^4}{4} + C_1 \textrm{ where } u = \ln{(y)} \implies \mathrm{d}u = \frac{1}{y}\,\mathrm{d}y \\ \frac{u^4}{4} + C_2 &= \frac{x^4}{4} + C_1 \\ \frac{u^4}{4} &= \frac{x^4}{4} + C_1 - C_2 \\ u^4 &= x^4 + 4\,C_1 - 4\,C_2 \\ \left[ \ln{(y)} \right] ^4 &= x^4 + C \textrm{ where } C = 4\,C_1 - 4\,C_2 \\ \ln{(y)} &= \pm \sqrt[4]{ x^4 + C } \\ y &= \mathrm{e}^{ \pm \sqrt[4]{x^4 + C} } \end{align*}$

Now since when $\displaystyle \begin{align*} x = 1, \, y = \mathrm{e}^2 \end{align*}$ we have

$\displaystyle \begin{align*} \mathrm{e}^2 &= \mathrm{e}^{ \pm \sqrt[4]{ 1^4 + C } } \\ 2 &= \pm \sqrt[4]{ 1 + C } \\ 16 &= 1 + C \\ C &= 15 \end{align*}$

and thus $\displaystyle \begin{align*} y = \mathrm{e}^{\pm \sqrt{ x^4 + 15 }} \end{align*}$.

 

[HallsofIvy]

Is there anywhere in the problem that says you are actually required to solve for y?