If y(x) is the solution of the differential equation dy \over dx + ( 2x + 1 \over x )y = e^ - 2x,\,\,x \gt 0,\, where y( 1 ) = 1 \over 2e^ - 2, then

y(x) is decreasing in ( 1 \over 2,1 )

Differential Equations PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q24 with Solution

View written solutionFree

Correct answer: D

Given differential equation

\frac{dy}{dx}+\left(\frac{2x+1}{x}\right)y=e^{-2x}, \qquad x>0

Since

\frac{2x+1}{x}=2+\frac{1}{x},

this is a linear differential equation:

\frac{dy}{dx}+\left(2+\frac{1}{x}\right)y=e^{-2x}.

Find the integrating factor

\text{I.F.}=e^{\int \left(2+\frac{1}{x}\right)dx}=e^{2x+\ln x}=xe^{2x}

(because $x>0$ ).

Multiplying the equation by $xe^{2x}$ :

xe^{2x}\frac{dy}{dx}+xe^{2x}\left(2+\frac{1}{x}\right)y = xe^{2x}e^{-2x}=x

The left side becomes:

\frac{d}{dx}\left(xe^{2x}y\right)=x

So,

xe^{2x}y=\int x\,dx=\frac{x^2}{2}+C

Hence,

y=\frac{\frac{x^2}{2}+C}{xe^{2x}}=e^{-2x}\left(\frac{x}{2}+\frac{C}{x}\right)

Use the initial condition

Given

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

Substitute $x=1$ :

e^{-2}\left(\frac{1}{2}+C\right)=\frac{1}{2}e^{-2}

Thus,

\frac{1}{2}+C=\frac{1}{2} \implies C=0

Therefore the solution is

\boxed{y(x)=\frac{x}{2}e^{-2x}}

Check options involving $y(\ln 2)$

y(\ln 2)=\frac{\ln 2}{2}e^{-2\ln 2}

Now,

e^{-2\ln 2}=e^{\ln(2^{-2})}=\frac{1}{4}

So,

y(\ln 2)=\frac{\ln 2}{2}\cdot \frac{1}{4}=\frac{\ln 2}{8}

Option A says $y(\ln 2)=\ln 4=2\ln 2$ , which is false.
Option C says $y(\ln 2)=\dfrac{\ln 2}{4}$ , which is false.

So A and C are incorrect.

Check monotonicity of $y(x)$

We have

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

Differentiate:

y'(x)=\frac{1}{2}e^{-2x}+\frac{x}{2}(-2)e^{-2x}

y'(x)=e^{-2x}\left(\frac{1}{2}-x\right)

Since $e^{-2x}>0$ for all $x$ , the sign of $y'(x)$ depends on $\frac{1}{2}-x$ .

If $x<\frac{1}{2}$ , then $y'(x)>0$ so $y$ is increasing.
If $x>\frac{1}{2}$ , then $y'(x)<0$ so $y$ is decreasing.

Therefore:

On $(0,1)$ , the function is not decreasing throughout, because it is increasing on $(0,\tfrac12)$ . So B is false.
On $\left(\tfrac12,1\right)$ , we have $y'(x)<0$ throughout. So D is true.

Final conclusion

The only correct option is:

\boxed{\text{D}}

This matches the stored correct answer.

More from Differential Equations

View written solutionFree

Correct answer: D

Given differential equation

\frac{dy}{dx}+\left(\frac{2x+1}{x}\right)y=e^{-2x}, \qquad x>0

Since

\frac{2x+1}{x}=2+\frac{1}{x},

this is a linear differential equation:

\frac{dy}{dx}+\left(2+\frac{1}{x}\right)y=e^{-2x}.

Find the integrating factor

\text{I.F.}=e^{\int \left(2+\frac{1}{x}\right)dx}=e^{2x+\ln x}=xe^{2x}

(because $x>0$ ).

Multiplying the equation by $xe^{2x}$ :

xe^{2x}\frac{dy}{dx}+xe^{2x}\left(2+\frac{1}{x}\right)y = xe^{2x}e^{-2x}=x

The left side becomes:

\frac{d}{dx}\left(xe^{2x}y\right)=x

So,

xe^{2x}y=\int x\,dx=\frac{x^2}{2}+C

Hence,

y=\frac{\frac{x^2}{2}+C}{xe^{2x}}=e^{-2x}\left(\frac{x}{2}+\frac{C}{x}\right)

Use the initial condition

Given

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

Substitute $x=1$ :

e^{-2}\left(\frac{1}{2}+C\right)=\frac{1}{2}e^{-2}

Thus,

\frac{1}{2}+C=\frac{1}{2} \implies C=0

Therefore the solution is

\boxed{y(x)=\frac{x}{2}e^{-2x}}

Check options involving $y(\ln 2)$

y(\ln 2)=\frac{\ln 2}{2}e^{-2\ln 2}

Now,

e^{-2\ln 2}=e^{\ln(2^{-2})}=\frac{1}{4}

So,

y(\ln 2)=\frac{\ln 2}{2}\cdot \frac{1}{4}=\frac{\ln 2}{8}

Option A says $y(\ln 2)=\ln 4=2\ln 2$ , which is false.
Option C says $y(\ln 2)=\dfrac{\ln 2}{4}$ , which is false.

So A and C are incorrect.

Check monotonicity of $y(x)$

We have

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

Differentiate:

y'(x)=\frac{1}{2}e^{-2x}+\frac{x}{2}(-2)e^{-2x}

y'(x)=e^{-2x}\left(\frac{1}{2}-x\right)

Since $e^{-2x}>0$ for all $x$ , the sign of $y'(x)$ depends on $\frac{1}{2}-x$ .

If $x<\frac{1}{2}$ , then $y'(x)>0$ so $y$ is increasing.
If $x>\frac{1}{2}$ , then $y'(x)<0$ so $y$ is decreasing.

Therefore:

On $(0,1)$ , the function is not decreasing throughout, because it is increasing on $(0,\tfrac12)$ . So B is false.
On $\left(\tfrac12,1\right)$ , we have $y'(x)<0$ throughout. So D is true.

Final conclusion

The only correct option is:

\boxed{\text{D}}

This matches the stored correct answer.

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