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

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

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Cristiano Sears

$$y'-\lambda y=0$$
$$y= Ce^{at}$$
$$\Rightarrow y'= Cae^{at}$$
Plugging the values of y and y'
$$\Rightarrow \frac{Cae^{at}}{\lambda} Ce^{at}=0$$
$$\Rightarrow Ce^{at}(a-\lambda)=0$$
$$e^{at} !=0, C!=0$$
$$\Rightarrow (a-\lambda)=0$$
$$\Rightarrow a=\lambda$$
$$y= Ce^{\lambda t}$$
Lets find the particular solution
Let $$y= b+ct$$
$$\Rightarrow y'=c$$
Plugging in equation $$y'-\lambda y= 1-\lambda t$$
$$\Rightarrow c-\lambda(b+ct)= 1-\lambda t$$
$$\Rightarrow c-\lambda b-\lambda ct= 1-\lambda t$$
$$\Rightarrow c-\lambda b=1$$, $$-\lambda c= -\lambda$$
Solving the second equation
$$-\lambda c= -\lambda$$
$$\Rightarrow -\lambda c+\lambda=0$$
$$\Rightarrow \lambda(-c+1)=0$$
$$\Rightarrow -c+1=0$$
$$\Rightarrow c=1$$
Plugging in 1-st equation
$$\Rightarrow 1-\lambda b=1$$
$$\Rightarrow -\lambda b=0$$
$$\Rightarrow b=0$$
The solution to the DE is
$$\Rightarrow y= Ce^{\lambda t}+t$$
Using the initial condition: $$y(0)=0$$
$$\Rightarrow y(0)= Ce^{\lambda 0}+0=0$$
$$\Rightarrow C\times1=0$$
$$\Rightarrow C=0$$
The final solution $$\Rightarrow y=0e^{\lambda t}+t$$
$$\Rightarrow y=t$$