Question

Consider the differential equationy′′+3y′−10y=0.(a) Find the general solution to this differential equation.(b) Now solve the initial-value problem corresponding to y(0)=2 and y′(0)=10.

Answer 1

Answer:

Y= 2e^(5t)

Step-by-step explanation:

Taking Laplace of the given differential equation:

s^2+3s-10=0

s^2+5s-2s-10=0

s(s+5)-2(s+5) =0

(s-2) (s+5) =0

s=2, s=-5

Hence, the general solution will be:

Y=Ae^(-2t)+ Be^(5t)………………………………(D)

Put t = 0 in equation (D)

Y (0) =A+B

2 =A+B……………………………………… (i)

Now take derivative of (D) with respect to "t", we get:

Y=-2Ae^(-2t)+5Be^(5t)   ....................... (E)

Put t = 0 in equation (E) we get:

Y’ (0) = -2A+5B

10  = -2A+5B ……………………………………(ii)

2(i) + (ii) =>

2A+2B=4 .....................(iii)

-2A+5B=10 .................(iv)

Solving (iii) and (iv)

7B=14

B=2

Now put B=2 in (i)

A=2-2

A=0

By putting the values of A and B in equation (D)

Y= 2e^(5t)