Question

find the initial value of \[y^2 dy/dt=y^2+ty-t^2\] . please help me

Answer 1

oh i forgot y(e)=0

Answer 2

Have you tried using y=vt?

Answer 3

i havent tried yet, im trying now

Answer 4

can you please show me how ? i've tried let y=vt and dy/dv=v+t dv/dt but still cant get it

Answer 5

From y=vt, you take dy/dt not dy/dv

Answer 6

opps sorry typo..

Answer 7

Sub it in
\[t^2v^2(v+t\frac{dv}{dt})=t^2v^2+t^2v-t^2\]
Simplify/Factor
\[t^2v^2(v+t\frac{dv}{dt})=t^2(v^2+v-1)\]
\[\frac{dv }{dt}=\frac{-v^3+v^2+v-1}{v^2t}\]

Answer 8

\[\int\limits \frac{v^2}{-v^3+v^2+v-1}dv=\int\limits \frac{1}{t}dt\]

Answer 9

There you go

Answer 10

omg, thank you so much

Answer 11

yw