1. find implicit form

hi
find in implicit form the general solution of differential equation

$\displaystyle \frac{dy}{dx}=3y^2e^{-2x}\sqrt{8+e^{-2x}}$

=$\displaystyle \int{3y^2}=\int{e^{-2x}}\sqrt{8+e^{-2x}}$

=$\displaystyle \int{3y^2}=\int{e^{-2x}}+e^{\frac{-3}{2}}+8^{\frac{1}{2}}$

$\displaystyle y^3=\frac{-1}{2}e^{-2x}-\frac{2}{3}e^{\frac{3}{2}}+\frac{2}{3}8x^\frac{3}{ 2}$

i did try but as you can see not sure what i am doing could do with some expert help please

2. Originally Posted by smartcar29
hi
find in implicit form the general solution of differential equation

$\displaystyle \frac{dy}{dx}=3y^2e^{-2x}\sqrt{8+e^{-2x}}$

=$\displaystyle \int{3y^2}=\int{e^{-2x}}\sqrt{8+e^{-2x}}$

=$\displaystyle \int{3y^2}=\int{e^{-2x}}+e^{\frac{-3}{2}}+8^{\frac{1}{2}}$

$\displaystyle y^3=\frac{-1}{2}e^{-2x}-\frac{2}{3}e^{\frac{3}{2}}+\frac{2}{3}8x^\frac{3}{ 2}$.

i did try but as you can see not sure what i am doing could do with some expert help please
When you separate the variables, you get $\displaystyle \frac{\,dy}{y^2}=3e^{-2x}\sqrt{8+e^{-2x}}\,dx$. So when you integrate the LHS, you get $\displaystyle -\frac{1}{y}+C$. When you integrate the RHS, you apply the substitution $\displaystyle u=8+e^{-2x}\implies \,du=-2e^{-2x}\,dx$. So it follows that $\displaystyle 3\int e^{-2x}\sqrt{8+e^{-2x}}\,dx=-\tfrac{3}{2}\int\sqrt{u}\,du=-u^{3/2}+C=-\left(8+e^{-2x}\right)^{3/2}+C$

So it follows now that $\displaystyle \frac{1}{y}+C=-\left(8+e^{-2x}\right)^{3/2}+C\implies \frac{1}{y}=-\left(8+e^{-2x}\right)^{3/2}+K$. This would be the implicit solution.

The explicit solution would be $\displaystyle y=\frac{1}{K-\left(8+e^{-2x}\right)^{3/2}}$

3. thank you for your much needed help i really suck at math, going to spend less time playing xbox and more time doing revision.

thanks again.