solving a linear equation

I finished this problem... but will you look at it and tell me if it's right, please?

Solve the linear equation (1+t)x' + x = cos(t) subject to x(0)=1.

By manipulating the equation I get:

(1+t)x'= -x +cos(t) and

x'=(-x + cos(t) )/(1+t) which is equal to
x'= -x/(1+t) +cos(t)/(1+t)
then... I manipulate it some more and get

x' + (1/(1+t))(x) = cos(t)/(1+t)

I then try to find the integrating factor
e^∫(1/(1+t)) and that is e^( ln (t+1))

So the integrating factor is t+1... I take that times the equation and get

[(t+1)x]'= cos(t)

So then I integrate both sides and get:
x(t+1)=sin(t) +c
x= sin(t)/(t+1) + c

c=1

x= sin(t)/(t+1) + 1


Is that right?

mathstresser said:

So then I integrate both sides and get:

x(t+1)=sin(t) +c
x= sin(t)/(t+1) + c

Click to expand...

Minor mistake there on the second equation...

\(\displaystyle \L x = \frac{\sin(t) + C}{t+1} \text{ \;\; not } \frac{\sin(t)}{t+1} + C\)

C turns out to be 1 again, but your x and the correct x are not the same!

Thank you!