find implicit form

smartcar29 · #1 $Old$ July 5th, 2009, 01:43 PM

hi
question reads
find in implicit form the general solution of differential equation

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

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

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

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

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

Chris L T521 · #2 $Old$ July 5th, 2009, 01:56 PM

Quote:

Originally Posted by smartcar29 $View Post$

hi
question reads
find in implicit form the general solution of differential equation

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

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

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

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

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 $\frac{\,dy}{y^2}=3e^{-2x}\sqrt{8+e^{-2x}}\,dx$ . So when you integrate the LHS, you get $-\frac{1}{y}+C$ . When you integrate the RHS, you apply the substitution $u=8+e^{-2x}\implies \,du=-2e^{-2x}\,dx$ . So it follows that $3\int e^{-2x}\sqrt{8+e^{-2x}}\,dx=-\tfrac{3}{2}\int\sqrt{u}\,du=-u^{3/2}+C=-\left(8+e^{-2x}\right)^{3/2}+C$

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

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

smartcar29 · #3 $Old$ July 5th, 2009, 02:08 PM

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.