Let y = y(x), x > 1, be the solution of the differential equation (x - 1)dy \over dx + 2xy = 1 \over x - 1 , with y(2) = 1 + e^4 \over 2e^4 . If y(3) = e^\alpha + 1 \over \beta e^\alpha , then the value of \alpha + \beta is equal to \underline\hspace2cm .

Differential Equations PYQ — JEE Main 2022 · 29 Jun · Shift 2 · Q36 with Solution

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Correct answer: 14

Rewrite the differential equation in linear form

Given

(x-1)\frac{dy}{dx}+2xy=\frac{1}{x-1}, \qquad x>1.

Divide throughout by $x-1$ :

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

This is a linear differential equation:

\frac{dy}{dx}+P(x)y=Q(x),

with

P(x)=\frac{2x}{x-1}, \qquad Q(x)=\frac{1}{(x-1)^2}.

Find the integrating factor

We compute

\int P(x)\,dx=\int \frac{2x}{x-1}\,dx.

Now,

\frac{2x}{x-1}=2+\frac{2}{x-1}.

So,

\int \frac{2x}{x-1}\,dx=\int \left(2+\frac{2}{x-1}\right)dx=2x+2\ln(x-1).

Hence the integrating factor is

\mathrm{IF}=e^{2x+2\ln(x-1)}=e^{2x}(x-1)^2,

(since $x>1$ , we need not use absolute value).

Multiply the equation by the integrating factor

Multiplying throughout by $e^{2x}(x-1)^2$ :

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

The left-hand side becomes

\frac{d}{dx}\left[y\,e^{2x}(x-1)^2\right].

Thus,

\frac{d}{dx}\left[y\,e^{2x}(x-1)^2\right]=e^{2x}.

Integrate

Integrating both sides,

y\,e^{2x}(x-1)^2=\int e^{2x}\,dx=\frac{1}{2}e^{2x}+C.

Therefore,

y=\frac{\frac12 e^{2x}+C}{e^{2x}(x-1)^2}.

Use the initial condition $y(2)=\dfrac{1+e^4}{2e^4}$

At $x=2$ , we have $(x-1)^2=1$ and $e^{2x}=e^4$ . So

y(2)=\frac{\frac12 e^4+C}{e^4}.

Given

\frac{\frac12 e^4+C}{e^4}=\frac{1+e^4}{2e^4}.

Multiply by $e^4$ :

\frac12 e^4+C=\frac{1+e^4}{2}=\frac12+\frac12 e^4.

Hence,

C=\frac12.

So the solution is

y=\frac{\frac12 e^{2x}+\frac12}{e^{2x}(x-1)^2} =\frac{e^{2x}+1}{2e^{2x}(x-1)^2}.

Compute $y(3)$

Substitute $x=3$ :

y(3)=\frac{e^6+1}{2e^6(3-1)^2}= rac{e^6+1}{2e^6\cdot 4}= rac{e^6+1}{8e^6}.

This matches the form

y(3)=\frac{e^\alpha+1}{\beta e^\alpha}.

So,

\alpha=6, \qquad \beta=8.

Therefore,

\alpha+\beta=6+8=14.

More from Differential Equations

View written solutionFree

Correct answer: 14

Rewrite the differential equation in linear form

Given

(x-1)\frac{dy}{dx}+2xy=\frac{1}{x-1}, \qquad x>1.

Divide throughout by $x-1$ :

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

This is a linear differential equation:

\frac{dy}{dx}+P(x)y=Q(x),

with

P(x)=\frac{2x}{x-1}, \qquad Q(x)=\frac{1}{(x-1)^2}.

Find the integrating factor

We compute

\int P(x)\,dx=\int \frac{2x}{x-1}\,dx.

Now,

\frac{2x}{x-1}=2+\frac{2}{x-1}.

So,

\int \frac{2x}{x-1}\,dx=\int \left(2+\frac{2}{x-1}\right)dx=2x+2\ln(x-1).

Hence the integrating factor is

\mathrm{IF}=e^{2x+2\ln(x-1)}=e^{2x}(x-1)^2,

(since $x>1$ , we need not use absolute value).

Multiply the equation by the integrating factor

Multiplying throughout by $e^{2x}(x-1)^2$ :

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

The left-hand side becomes

\frac{d}{dx}\left[y\,e^{2x}(x-1)^2\right].

Thus,

\frac{d}{dx}\left[y\,e^{2x}(x-1)^2\right]=e^{2x}.

Integrate

Integrating both sides,

y\,e^{2x}(x-1)^2=\int e^{2x}\,dx=\frac{1}{2}e^{2x}+C.

Therefore,

y=\frac{\frac12 e^{2x}+C}{e^{2x}(x-1)^2}.

Use the initial condition $y(2)=\dfrac{1+e^4}{2e^4}$

At $x=2$ , we have $(x-1)^2=1$ and $e^{2x}=e^4$ . So

y(2)=\frac{\frac12 e^4+C}{e^4}.

Given

\frac{\frac12 e^4+C}{e^4}=\frac{1+e^4}{2e^4}.

Multiply by $e^4$ :

\frac12 e^4+C=\frac{1+e^4}{2}=\frac12+\frac12 e^4.

Hence,

C=\frac12.

So the solution is

y=\frac{\frac12 e^{2x}+\frac12}{e^{2x}(x-1)^2} =\frac{e^{2x}+1}{2e^{2x}(x-1)^2}.

Compute $y(3)$

Substitute $x=3$ :

y(3)=\frac{e^6+1}{2e^6(3-1)^2}= rac{e^6+1}{2e^6\cdot 4}= rac{e^6+1}{8e^6}.

This matches the form

y(3)=\frac{e^\alpha+1}{\beta e^\alpha}.

So,

\alpha=6, \qquad \beta=8.

Therefore,

\alpha+\beta=6+8=14.

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