Question

Solve IVP
t^2 dy/dt = 2cos^2 t , y(0)=pi/4

Answer 1

\[t ^{2}\frac{ dy }{ dt }=2 \cos ^{2}t \]
\[y(0)=\frac{ \Pi }{ 4 }\]

Answer 2

no solution
y(t) for all t>0 equals infinity since y(0)=pi/4 and y'(0)=infinity

Answer 3

i'm confused sorry.

Answer 4

gomen

Answer 5

maybe the equation is supposed to be t^2 dy/dt = 2sin^2 t

Answer 6

otherwise, at t=0,
t^2 dy/dt = 2cos^2 t
0 * dy/dt = 2
dy/dt=2/0=infinity

Answer 7

http://www.wolframalpha.com/input/?i=x%5E2+dy%2Fdx%3D2cos%5E2+x

Answer 8

that only works if the initial condition is y(0)=-infinity

Answer 9

i mainly need help with the integration part

Answer 10

dy/dt=2cos^2 t/t^2
this can be expressed in terms of the sinintegral function Si(x)=integrate(sin(x)/x)

Answer 11

one way is to reduce cos^2 t down to cos(2t) using the double angle identity, then use integration by parts to reduce the t^2 down to t

Answer 12

that makes sense so then that gets rid of the 2 as well