y' = y(sin t) + 3 Is /sci/ good enough at undergrad

>>7725894
Solve what

>>7725894
yipe. that's difficult. You mean
y'(t) = y(sin(t)) + 3 ?

>>7725894
do you mean y(sin t) as in function y of sin t, or y times sin t?

[math]y(t)=3 e^{-cos(t)} \left[ \int e^{cos(t)} dt - c\right] [/math]

What Am I supposed to solve? second y''(t), y(t) or y'(t) after inputting y(t)?

>>7725942
No... He means
y'(t) = y(t)*sin(t) +3
Solve the diffeq faggots.
y'(t) - y(t)sin(t) = 3
[math]
I(t) = e^{\int sin(t) dt} = e^{-cos(t)}
[/math]
[math]
(I(t)y)' = 3e^{-cos(t)}
[/math]
[math]
y = c + 3e^{cost} * \int e^{-cos(t) }
[/math]
like other anon said

>>7725894
Solve
[math]
y' - xy = x^2
[/math]
by power series :^)

>>7725998
Oh, almost forgot y(0) = 0

>>7725894
dy/dt = y(sin t) +3
dy/dsint dsint/dt = y(sin t) +3
y'(sin(t))cos(t)= y(sin(t)) +3
y'(sin(t))cos(t) - y(sin(t)) = 3
y'(sin(t)) - y(sin(t))*sec(t) = -3sec(t)
d/dsin(t)[y(sin(t))e^tan(t)] e^-tan(t) = -3sec(t)
y(sin(t))e^tan(t) = ∫ -3sec(t)e^tan(t) dsin(t)
y(sin(t)) = e^-tan(t) ∫ -3sec^2(t) e^tan(t) dt
y(sin(t)) = e^-tan(t) ∫ -3e^u du
y(sin(t)) = e^-tan(t)*[-3e^tan(t)+C]= -3 +Ce^-tan(t)
y(x)= -3 + Ce^-[x/√(1-x^2)]

>>7726037
>dy/dt = y(sin t) +3
>dy/dsint dsint/dt = y(sin t) +3
>y'(sin(t))cos(t)= y(sin(t)) +3

x=sin(t)
y'(x)*cos(sin^-1(x))=y(x)+3
y'(x)√(1-x^2)=y(x)+3
y'(x)-1/√(1-x^2) y(x) = 3/√(1-x^2)
d/dx[y(x)e^-sin^-1(x)]e^sin^-1(x) = 3/√(1-x^2)
y(x)e^-sin^-1(x)= -3e^sin^-1(x) + C
y(x) = -3 + Ce^sin^-1(x)
y(t) = -3 + Ce^t