Expert Community at Your Service
Solve your problem for the price of one coffee
Step 1 L{e3tsin(4t)−t4+et} =L{e3tsin(4t)}−L{t4}+L{et} 1) we know that L{tn}=n!sn+1L{eat}=1s−a L{sin(at)}=as2+a2 L{sin(4t)}=4s2+16: by first shifting L{eatsin(at)}=b(s−a)2+b2 L{e3tsin(4t)}=4(s−3)2+16, L{t4}=4!s4+1=24s5 L{et}=1s−1 Step 2 Then 1) will becomes as L{e3tsin(t)−t4+et}=4(s−3)2+16−24s5+1s−1
Answer is given below (on video)
Ask your question. Get your answer. Easy as that
Solve the initial value problem using Laplace transforms.y"+y=f(t),y(0)=0,y′(0)=1Heref(t)={00≤t<3π1t≥3π
Get answers within minutes and finish your homework faster
Or
Dont have an account? Register
Create a free account to see answers
Already have an account? Sign in