What is the solution of dy/dx+2y=1 Satisfying y(0)=0 ?

Correct Answer is : y=1−e−2x2y={1-e^{-2x}}{2}y=21−e−2x

Correct Answer is : y=1−e−2x2y={1-e^{-2x}}{2}y=21−e−2x

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UPSC NDA Math Model Paper 3

Question : 60 of 120

Marks: +1, -0

\frac{dy}{dx}+2y=1

y(0)=0 ?

Solution:

Given:

\frac{dy}{dx}+2y=1

and

y(0)=0

So we have,

\frac{dy}{dx}+2y=1

\Rightarrow\frac{dy}{dx}=1-2y

\Rightarrow\frac{dy}{1-2y}=dx

On integrating both sides,

\Rightarrow\int\frac{dy}{1-2y}=\int dx

\Rightarrow-\frac{\ln(1-2y)}{2}=x+C

\Rightarrow\frac{\ln(1-2y)}{2}+x+C=0

Now, we know at

x=0,\ y=0.

So, putting values of

x

and

y

we get,

\Rightarrow\frac{\ln(1-0)}{2}=0+C

\Rightarrow C=0

So, the equation becomes,

\Rightarrow\frac{\ln(1-2y)}{2}+x=0

\Rightarrow\ln(1-2y)=-2x

\Rightarrow(1-2y)=e^{-2x}

\Rightarrow 2y=1-e^{-2x}

\Rightarrow y=\frac{1-e^{-2x}}{2}

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