Use laplace transforms to solve y’’(t) + 4y(t) = ф(t) with y(0) = 1, y’(0) = 0, where

ф(t) = ((4t,0≤t≤1 and 4, t1))

Question

Use laplace transforms to solve y’’(t) + 4y(t) = ф(t) with y(0) = 1, y’(0) = 0, where

ф(t) = ((4t,0≤t≤1 and 4, t>1))

anonymous · Answer

you have no idea how we could do this ?

anonymous · Answer

lol! no! can you give me a start?

anonymous · Answer

ok the linearity of the laplace transform can help ! 
L(y"+4y)=L(phi(t))

anonymous · Answer

still not having the answer ?

anonymous · Answer

haha! no...

anonymous · Answer

I will write the answer step by step...when you will be able to continue by yourself...tell me to stop kk ?

anonymous · Answer

sweet! u will be my lifesaver!!!

anonymous · Answer

$$L(y"+4y)=L(\phi)$$
then $$L(y")+L(4y)=L(\phi)$$

anonymous · Answer

I guess you can compute L(phi) right .?

anonymous · Answer

humm...use the formula your just wrote?

anonymous · Answer

$$L(y")=s^2L(y)-sy'(0)-y(0)$$ and $$L(4y)=4L(y)$$
so,the equation becomes $$s^2L(y)-1+4L(y)=L(\phi)$$

anonymous · Answer

still with me .?

anonymous · Answer

nice...how about this...v tried to solve it...can u check it and see if there are any errors? give me 10 min

anonymous · Answer

okk that sounds great ! you can send a message to me... if you want to !

anonymous · Answer

cool!

anonymous · Answer

I dont see how to attach a file in msg! so...here it is!

anonymous · Answer

for computing L(phi) you can also use the definition...it's a lot easier :) but you did seems correct ! 
w8 5 more minutes...to check if there is a miscalculation

anonymous · Answer

ok take your time!

anonymous · Answer

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