Solve \frac{dx}{dt}+2x=2e^{-3t} with initial condition x(0)=2

Yulia 2021-09-09 Answered

Solve dxdt+2x=2e3t with initial condition x(0)=2

You can still ask an expert for help

Want to know more about Laplace transform?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

SabadisO
Answered 2021-09-10 Author has 108 answers

Step 1
Consider the provided differential equation,
Given dxdt+2x=2e3t with initial condition x(0)=2
This is the ordinary differential equation.
Take Laplace on both the sides,
L(dxdt+2x)=L(2e3t)
L(dxdt)+2L(x)=2L(e3t)
sLx(t)x(0)+2Lx(t)=2s+3[Since ,  L(eat)=1s+a]
sL(x(t))2+2L(x(t))=2s+3
L(x(t))(s+2)=2s+3+2
L(x(t))(s+2)=2+2(s+3)s+3
L(x(t))(s+2)=2s+8s+3
Step 2
Now, further simplified the above equation,
L(x(t))=2s+8(s+3)(s+2)
L(x(t))=2(s+4)(s+3)(s+2)
L(x(t))=2[2(s+3)(s2)](s+3)(s+2)
L(x(t))=4(s+3)2(s+2)(s+3)(s+2)
L(x(t))=4(s+2)2(s+3)
Now ,

We have step-by-step solutions for your answer!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

New questions