Question

Find the solution of the following nonhomogeneous 2nd order linear initial value problem: | 1. y” + 7y + 10y = 176e6t, y (0) = 0, y'(0) = 13 2. y” + 7y + 10y = 140 cos(4t) – 30 sin(4t) y(0) = 1, y'(0) = 0

Answer 1

1. Given differential equation ,

y" + 7y' + 10y = 176c6ty(0) = 0.7(0) = 13

First we will find complementary part of the solution that is solution of .

y" + 7y' + 10y = 0

Let $y=e^{mt}$ be a solution of .

y" + 7y' + 10y = 0

Now , $y=e^{mt}$

→y = mety" = mmt

Using these values in we get ,

y" + 7y' + 10y = 0

memt + 7met +10emt = 0

→ (m + 7m + 10) = 0

→ m² + 7m + 10 = 0
, since
07 que
.

= m+5)(m + 2) = 0

→ m= -5.-2

So the required complementary solution is ,

y..x) = Ce-5t + ce-2
, where $c_{1},c_{2}$ are arbitrary constant .

Now we will particular solution ,

D2 + 7D + 10 (1766)

_1761 = 176 D2 + 7D + 10

= 176 62 +7,6 + 10

= 1766 36 +42 + 10

= 1766 1

= 261

So the solution of the given differential equation is ,

ya) = ce-5t + coe-2 +2e6t

Now ,

y(0) = 0

→ C + C + 2 = 0

→ C+C = -2
  

Also ,

y (0) = 13

☆ -5c1 - 202 +12 = 13

→ -5c1 – 2c2 = 1
  

gives ,

2 x (i) + (i)

-3c1 = -3

c=1

Substituting value of $c_{1}$ in (i) we get ,

1+c= -2

→ = -3

Sunstituting values of $c_{1},c_{2}$ the required solution is ,

y(x) = e-bt _ 3 e 2+ + 2e6t

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If you have doubt or need more clarification at any step please comment .