Solve the differential equation y'-lambda y= 1-lambda t, y(0)=0

Question
Solve the differential equation \(y'-lambda y= 1-lambda t\), y(0)=0

Answers (1)

2021-03-09
\(y'-lambda y=0\)
\(y= Ce^(at)\)
\(=> y'= Cae^(at)\)
Plugging the values of y and y'
\(=> Cae^(at)/ lambda Ce^(at)=0\)
\(=> Ce^(at)(a-lambda)=0\)
\(e^(at) !=0, C!=0\)
\(=> (a-lambda)=0\)
\(=> a=lambda\)
\(y= Ce^(lambda t)\)
Lets find the particular solution
Let \(y= b+ct\)
\(=> y'=c\)
Plugging in equation \(y'-lambda y= 1-lambda t\)
\(=> c-lambda(b+ct)= 1-lambda t\)
\(=> c-lambda b-lambda ct= 1-lambda t\)
\(=> c-lambda b=1\), \(-lambda c= -lambda\)
Solving the second equation
\(-lambda c= -lambda\)
\(=> -lambda c+lambda=0\)
\(=> lambda(-c+1)=0\)
\(=> -c+1=0\)
\(=> c=1\)
Plugging in 1-st equation
\(=> 1-lambda b=1\)
\(=> -lambda b=0\)
\(=> b=0\)
The solution to the DE is
\(=> y= Ce^(lambda t)+t\)
Using the initial condition: y(0)=0
\(=> y(0)= Ce^(lambda 0)+0=0\)
\(=> C*1=0\)
\(=> C=0\)
The final solution \(=> y=0e^(lambda t)+t\)
=> y=t
0

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