Question

Solve the given initial value problem. y'' - 10y' + 25y = 0; y(0) = -3, y'(0) = 57 4 The solution is y(t) =

thank you!!

Answer 1

Here I have solved the problem considering as a function of
T
, i.e., .

But in the problem it is given that i.e., as a function of .

The solution will remain exactly same. Just replace each
T
with in the following solution.

Initial Value Problem: Y" - 10 y' + 25y = 0 Initial Conditions yco) = – 3 (2) y'(o)= 57 4 The corresponding auxiliary equation is k ² – 10k + 25 = 0 (3) ► 2x5k + 52 = 0 (k - 5)² = 0 k = 5,5 The general solution of the ordinary differential equation (1) is given by y(x) = (C, + (₂x) e 5 x where and C2 are arbitrary constants, For the initial value problem with the initial conditions (2)

have to find the constants e & C2 Before that, note Y'(x) = {50, + (5 x +1) ez} 5 x (using product rule of differention and re-arraging ) Now from (2) (4) , yo) - 3 nie., 5xo (6, + C2 xo) e - - 3 i.eu c = -3 len y'co) - 5 57 4 ine, {5c, + (5 x + 1) c₂} 5 xo e 11 - 57 4 ine, 5 c, & c. U - 57 А C2 - 57 5*(-3) (by (5)) 3 4. . The of the given solution is initial value 5 x problem y(x) = (-3 + x) e