# 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

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

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