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

First order differential equations
asked 2021-03-08

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

Answers (1)


\(y'-\lambda y=0\)
\(y= Ce^{at}\)
\(\Rightarrow y'= Cae^{at}\)
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 \(\Rightarrow y=0e^{\lambda t}+t\)
\(\Rightarrow y=t\)

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