Solving 10y′ + 10y = 7exp(6x) with initial condition y(0) = 7

[ganondorf29]

Homework Statement

10 y′+ 10 y = 7 exp(6*x)
y ( 0 ) = 7

Homework Equations

The Attempt at a Solution

First I divided everything by 10 because the coefficient of the y' should be 1
Then I found the integrating factor, which was exp(integral of 1*dx) = exp(x) = Integrating factor

I multiplied to all three terms:
exp(x)*y' + exp(x)*y = (7*exp(2x))/10
(y*exp(x))' = (7*exp(2x))/10

I then integrated the above and got
y*exp(x) = exp(7x)/10 + C
y = exp(6x)/10 + C

Solving for c y ( 0 ) = 7

7 = exp(0)/10 + C
7 = 1/10 + C
C = 69/10

I got my final answer to be y = (exp(6x)+69)/10

However, the my homework says that's wrong. Does anyone know where I messed up?

Thanks

 

[djeitnstine]

Your algebra is off, because when you divide the right side you should get y=exp(6x)/10 + C*exp(-x)

 

[ganondorf29]

Your algebra is off, because when you divide the right side you should get y=exp(6x)/10 + C*exp(-x)

I just realized that too. Thank you