Asked by Ella
Solve the IVP
{y'=-(sinx)y + xexp(cosx)
y(0) = 1
Please show me step by step. Do I need to find out if this is exact? Do I use an integrating factor?
{y'=-(sinx)y + xexp(cosx)
y(0) = 1
Please show me step by step. Do I need to find out if this is exact? Do I use an integrating factor?
Answers
Answered by
Count Iblis
Solve the homogeneous equation first:
y_h' = - sin(x) y_h ---->
y_h(x) = K exp[cos(x)]
Then, to find the solution, you use the variation of the constant method, i.e. take the solution of the homogeneous solution y_h and there you replace K by an unknown function K(x):
y(x) = K(x) exp[cos(x)]
If you substitute this in the differential equation then what happens is that only the term proportional to K' survives. Due to the product rule the terms proportional to K are what you would get if K were a constant, but these terms will satisfy the homogeneous equation, so they will sum to zero.
So, what you get is:
K' Exp[cos(x)] = x Exp[cos(x)] ----->
K = 1/2 x^2 + c ----->
y(x) = (1/2 x^2 + c)Exp[cos(x)]
y_h' = - sin(x) y_h ---->
y_h(x) = K exp[cos(x)]
Then, to find the solution, you use the variation of the constant method, i.e. take the solution of the homogeneous solution y_h and there you replace K by an unknown function K(x):
y(x) = K(x) exp[cos(x)]
If you substitute this in the differential equation then what happens is that only the term proportional to K' survives. Due to the product rule the terms proportional to K are what you would get if K were a constant, but these terms will satisfy the homogeneous equation, so they will sum to zero.
So, what you get is:
K' Exp[cos(x)] = x Exp[cos(x)] ----->
K = 1/2 x^2 + c ----->
y(x) = (1/2 x^2 + c)Exp[cos(x)]
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.