modeling w/differential equations

[Guest]

Find the solution to the differential equation y'=(x^2)y that satisfies the initial condition y(0)= 8.

I'm not really sure where to start.

 

[skeeter]

this DE is separable ...

dy/dx = x²y

dy/y = x² dx

do the integration.

 

[Guest]

Ok, so:

dy/y = x^2 dx
int (1/y) = int (x^2)
ln(y) = (x^3)/3 + c
y = e^(x^3/3 + c)
y = e^(x^3/3) * e^c
y = ce^(x^3/3)

then i substitute y=8 and x=0

8 = ce^0
8 = c * 1
c = 8

so the final solution is:
y = 8e^(x^3/3)

is that correct?

 

[galactus]

Sub it into your DE and see.

\(\displaystyle \L\\y=8e^{\frac{x^{3}}{3}}\)

\(\displaystyle \L\\y'=8x^{2}e^{\frac{x^{3}}{3}}\)

\(\displaystyle \H\\y'=x^{2}\underbrace{y}_{\searrow}\)
\(\displaystyle \L\\\overbrace{8x^{2}e^{\frac{x^{3}}{3}}}^{\nwarrow}=x^{2}(8e^{\frac{x^{3}}{3}})\)

YEP!!. Looks correct.

 

[Guest]

thank you