First Order Linear Differential Equation

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SUMMARY

The discussion focuses on solving the first-order linear differential equation given by 10y' + 10y = 7exp(6x) with the initial condition y(0) = 7. The correct approach involves dividing the entire equation by 10 to simplify the coefficient of y' to 1, finding the integrating factor exp(x), and applying it to both sides. The final solution is y = (exp(6x) + C*exp(-x))/10, where C is determined to be 69/10 after applying the initial condition. The error identified was in the algebraic manipulation of the right side of the equation.

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ganondorf29
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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
 
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Your algebra is off, because when you divide the right side you should get y=exp(6x)/10 + C*exp(-x)
 
djeitnstine said:
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
 

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