dy/dx= (5x^2)/sqrt(y)
and y=1 when x=0
Use separation of variables to solve the initial value problem

Question

dy/dx= (5x^2)/sqrt(y)
and y=1 when x=0
Use separation of variables to solve the initial value problem

zepdrix · Answer

$$\Large\bf\sf \frac{dy}{dx}=\frac{5x^2}{\sqrt y},\qquad y(0)=1$$Ok what are you stuck on? :O

anonymous · Answer

ok when i do it i got y=((5x^3)/2)^3/2

anonymous · Answer

is that correct?

zepdrix · Answer

Hmm let's check.

zepdrix · Answer

$$\Large\bf\sf \int\limits \sqrt y \;dy\quad=\quad \int\limits 5x^2\;dx$$Gives us,$$\Large\bf\sf \frac{2}{3}y^{3/2}\quad=\quad \frac{5}{3}x^3+C$$

zepdrix · Answer

Multiply both sides by 2/3,$$\Large\bf\sf y^{3/2}\quad=\quad \frac{5}{2}x^3+c$$

zepdrix · Answer

Plug in our initial data,$$\Large\bf\sf 1\quad=\quad \frac{5}{2}\cdot 0+c\qquad\implies\qquad c=1$$

zepdrix · Answer

Did you get your c value messed up maybe? Hmm

anonymous · Answer

i think that might be

anonymous · Answer

is it |dw:1392614228491:dw|