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## Answered question

2021-03-08

Solve the differential equation ${y}^{\prime }-\lambda y=1-\lambda t$,  $y\left(0\right)=0$

### Answer & Explanation

Cristiano Sears

Skilled2021-03-09Added 96 answers

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

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