given in series circuit L=1H, R=110ohm, C=0.001F, v=90V. Find the resultant current when the switch is originally open and it is closed for 1s. then it is open again when t=1. use laplace transform...
given in series circuit L=1H, R=110ohm, C=0.001F, v=90V. Find the resultant current when the switch is originally open and it is closed for 1s. then it is open again when t=1. use laplace transform...
Use the Laplace transform to solve y^(4)-4y'''+6y"-4y'+y=0; y(0)=0, y'(0)=1, y"(0)=0, y'''(0)=1. Answer: y=te^t-t^2*e^t+(2/3)t^3*e^t L(y^(4))-4L(y''')+6L(y")-4L(y')+L(y)=L(0) But...
Use the Laplace transform to solve y"-y'-6y=0; y(0)=1, y'(0)=-1. Answer: y=(1/5)(e^(3t)+4e^(-2t)) I don't know how to take the Laplace transform for both sides. Help me step by step...
Find the inverse Laplace transform of F(s)=G(s)/(s^2+1) by using the convolution theorem.