Question

solve the initial value problem tx''-x'=t^2, x (0)=0

Answer 1

Notice
\[
x_p(t)= \frac 1 3 t^3
\]
is a particular solution of the HE

Answer 2

Sorry
is a particular solution of the NHE

Answer 3

Find r so that
\[
x(t) = t^r
\]
is a solution to HE

Answer 4

You find r=0 and r =2
So the general equation of the HE is of the
form
\[
a + b t^2
\]
So the solution is of the form
\[

x(t)=a + b t^2+ \frac {t^3}3
\]

Answer 5

since x(0) =0, then a=0;
so
\[
b t^2 +\frac {t^3}3
\]

Answer 6

I meant
\[
x(t)=b t^2 +\frac {t^3}3
\]

Answer 7

You need another initial condition to find b

Answer 8

How do i solve this using laplace transform?

Answer 9

You're missing at least one other initial condition...

Answer 10

@SithsAndGiggles , nope. that's the way the equation is set up.