If the solution curve of the differential equation (y-2 \log _e x) d x+(x \log _e x^2) d y=0, x \gt 1 passes through the points (e, \frac43) and (e^4, \alpha) , then \alpha is equal to \underline\hspace2cm .

Differential Equations PYQ — JEE Main 2023 · 8 Apr · Shift 1 · Q41 with Solution

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Correct answer: 3

Write the differential equation in standard form

Given

\left(y-2\ln x\right)dx+\left(x\ln x^2\right)dy=0, \qquad x>1

Since $x>1$ , we have $\ln x^2 = 2\ln x$ . So the equation becomes

(y-2\ln x)dx + 2x\ln x\, dy = 0.

Hence,

2x\ln x\,\frac{dy}{dx} + y - 2\ln x = 0.

So,

\frac{dy}{dx} + \frac{1}{2x\ln x}y = \frac{1}{x}.

This is a linear differential equation.

Find the integrating factor

Here,

P(x)=\frac{1}{2x\ln x}.

So the integrating factor is

\text{I.F.} = e^{\int \frac{1}{2x\ln x}dx}.

Using $\int \frac{1}{x\ln x}dx = \ln(\ln x)$ ,

\text{I.F.} = e^{\frac12 \ln(\ln x)} = (\ln x)^{1/2}.

Solve the differential equation

Multiplying the equation by $(\ln x)^{1/2}$ ,

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

So,

\frac{d}{dx}\left(y\sqrt{\ln x}\right)=\frac{\sqrt{\ln x}}{x}.

Integrate both sides:

y\sqrt{\ln x} = \int \frac{\sqrt{\ln x}}{x}dx + C.

Let

t=\ln x \quad \Rightarrow \quad dt=\frac{dx}{x}.

Then

\int \frac{\sqrt{\ln x}}{x}dx = \int t^{1/2}dt = \frac{2}{3}t^{3/2} = \frac{2}{3}(\ln x)^{3/2}.

Thus,

y\sqrt{\ln x} = \frac{2}{3}(\ln x)^{3/2}+C.

Divide by $\sqrt{\ln x}$ :

y = \frac{2}{3}\ln x + \frac{C}{\sqrt{\ln x}}.

Use the point $\left(e,\frac{4}{3}\right)$

At $x=e$ , we have $\ln e=1$ . Hence

\frac{4}{3} = \frac{2}{3}(1) + C.

So,

C=\frac{2}{3}.

Therefore the solution curve is

y = \frac{2}{3}\ln x + \frac{2}{3\sqrt{\ln x}}.

Find $\alpha$ at $x=e^4$

For $x=e^4$ ,

\ln(e^4)=4,

\alpha = \frac{2}{3}\cdot 4 + \frac{2}{3\sqrt{4}} = \frac{8}{3} + \frac{2}{6} = \frac{8}{3} + \frac{1}{3} = 3.

Thus,

\boxed{\alpha=3}.

Compare with stored answer

Stored correct answer: $3$

Our derived answer is also $3$ , so it agrees.

More from Differential Equations

View written solutionFree

Correct answer: 3

Write the differential equation in standard form

Given

\left(y-2\ln x\right)dx+\left(x\ln x^2\right)dy=0, \qquad x>1

Since $x>1$ , we have $\ln x^2 = 2\ln x$ . So the equation becomes

(y-2\ln x)dx + 2x\ln x\, dy = 0.

Hence,

2x\ln x\,\frac{dy}{dx} + y - 2\ln x = 0.

So,

\frac{dy}{dx} + \frac{1}{2x\ln x}y = \frac{1}{x}.

This is a linear differential equation.

Find the integrating factor

Here,

P(x)=\frac{1}{2x\ln x}.

So the integrating factor is

\text{I.F.} = e^{\int \frac{1}{2x\ln x}dx}.

Using $\int \frac{1}{x\ln x}dx = \ln(\ln x)$ ,

\text{I.F.} = e^{\frac12 \ln(\ln x)} = (\ln x)^{1/2}.

Solve the differential equation

Multiplying the equation by $(\ln x)^{1/2}$ ,

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

So,

\frac{d}{dx}\left(y\sqrt{\ln x}\right)=\frac{\sqrt{\ln x}}{x}.

Integrate both sides:

y\sqrt{\ln x} = \int \frac{\sqrt{\ln x}}{x}dx + C.

Let

t=\ln x \quad \Rightarrow \quad dt=\frac{dx}{x}.

Then

\int \frac{\sqrt{\ln x}}{x}dx = \int t^{1/2}dt = \frac{2}{3}t^{3/2} = \frac{2}{3}(\ln x)^{3/2}.

Thus,

y\sqrt{\ln x} = \frac{2}{3}(\ln x)^{3/2}+C.

Divide by $\sqrt{\ln x}$ :

y = \frac{2}{3}\ln x + \frac{C}{\sqrt{\ln x}}.

Use the point $\left(e,\frac{4}{3}\right)$

At $x=e$ , we have $\ln e=1$ . Hence

\frac{4}{3} = \frac{2}{3}(1) + C.

So,

C=\frac{2}{3}.

Therefore the solution curve is

y = \frac{2}{3}\ln x + \frac{2}{3\sqrt{\ln x}}.

Find $\alpha$ at $x=e^4$

For $x=e^4$ ,

\ln(e^4)=4,

\alpha = \frac{2}{3}\cdot 4 + \frac{2}{3\sqrt{4}} = \frac{8}{3} + \frac{2}{6} = \frac{8}{3} + \frac{1}{3} = 3.

Thus,

\boxed{\alpha=3}.

Compare with stored answer

Stored correct answer: $3$

Our derived answer is also $3$ , so it agrees.

Differential Equations question

Differential Equations question

More from Differential Equations

Differential Equations question

Differential Equations question

More from Differential Equations