Asked by Sarah
Solve the differential equation y'=3t^2+4. Solve the initial value problem y(0)=3.
Separation of variables!
My work:
dy/dt= 3t^2+4
dy= 3t^2+4 dt
Then you integrate both sides.
∫ dy= ∫ 3t^2+4dt
Question: is there a 1 in dy? ( ∫ 1dy)? If so:
y+C1=t^3+4t+C2
y=t^3+4t+C (where c is constant)
Then you find C
y(0)=3
3= 0+0 +C , So C=3
Ans: y=t^3+4t+3
Is this correct? Thank you!
Separation of variables!
My work:
dy/dt= 3t^2+4
dy= 3t^2+4 dt
Then you integrate both sides.
∫ dy= ∫ 3t^2+4dt
Question: is there a 1 in dy? ( ∫ 1dy)? If so:
y+C1=t^3+4t+C2
y=t^3+4t+C (where c is constant)
Then you find C
y(0)=3
3= 0+0 +C , So C=3
Ans: y=t^3+4t+3
Is this correct? Thank you!
Answers
Answered by
Reiny
I agree