# calc problem differencial equation

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• Feb 26th 2007, 05:38 PM
clockingly
calc problem differencial equation
hi, i was wondering if i did this problem right. i had to solve the following differential equation for y as a function of t using separation of variables:

y' = yt^2

so this is what i did.

i set dy/dt = yt^2

then from that i got dy/y = t^2 dt, which is the same as 1/y dy = t^2 dt.

i took the integral of both sides.

the integral of 1/y dy is ln(y)

the integral of t^2 dt is t^3/3 + C

so then i got that y = e^(ln(t^3/3)+C)

this is the same as y = e^(ln(t^3/3) * e^C

then i got y = (t^3 * e^c)/3 for a final answer.
• Feb 26th 2007, 06:47 PM
ThePerfectHacker
Quote:

Originally Posted by clockingly
hi, i was wondering if i did this problem right. i had to solve the following differential equation for y as a function of t using separation of variables:

y' = yt^2

y=0 is a trivial solution.

Assume y not = 0 (in fact it is zero everywhere or nowhere).
Now divide,
y'/y=t^2
Thus,
INT y'/y dt = INT t^2 dt
ln |y| = (1/3)t^3+C
y=exp[(1/3)t^3 +C]=e^c * exp[1/3t^3]=C*e^{1/3t^3} for C>0.

Note, y=0 was also a solution. Corresponding for C=0.
Thus,
y=C*e^{1/3t^3} for C>=0.
Is the solution on this non-empty open interval.