The solution curve y = y(x) of the differential equation (1 + y^2)(1 + (x)) dx + x dy = 0, x 0, passes through the point (1, 1). If y(e) = - ( 32) + ( 32), then + 2 is:

Given: (1 + y^2)(1 + e x) dx + x dy = 0 with x 0, and the curve passes through (1,1). Find: + 2 if y(e) = - ( 32) + ( 32) Rewrite the differential equation in separable form: dydx = - (1+y^2)(1+ e x)x So, dy1+y^2 = - ( 1x + xx)dx Integrate both sides: dy1+y^2 = - ( 1x + xx)dx Hence, ^-1y = - x - ( x)^22 + C which is equivalently written as x + ( x)^22 + ^-1y = C Use the condition (x,y)=(1,1): 1 + ( 1)^22 + ^-1(1) = C 0 + 0 + 4 = C So, x + ( x)^22 + ^-1y = 4 Now put x=e: e + ( e)^22 + ^-1y(e) = 4 1 + 12 + ^-1y(e) = 4 Therefore, ^-1y(e) = 4 - 32 so y(e) = ( 4 - 32) Using (A-B) = A - B1 + A B with A= 4 and B= 32, y(e) = 1 - ( 32)1 + ( 32) Comparing with - ( 32) + ( 32) we get =1 and =1. Thus, + 2 = 1 + 2(1) = 3 Therefore, the correct option is B. Take v = 1 + e x Then dv = dxx The differential equation becomes dy1+y^2 = -v dv Integrating, dy1+y^2 = -v dv ^-1y = - v^22 + C Substitute back v = 1 + e x: ^-1y = - (1+ e x)^22 + C At (1,1), ^-1(1) = - (1+ e 1)^22 + C 4 = - 12 + C So, C = 4 + 12 Hence, ^-1y = - (1+ e x)^22 + 4 + 12 For x=e, we have 1+ e e = 2. Thus, ^-1y(e) = - 42 + 4 + 12 = 4 - 32 So again, y(e) = 1 - ( 32)1 + ( 32) Therefore =1, =1, and +2 =3.

The solution curve y = y(x) of the differential equation (1 + y^2… — Solution | JEE Mathematics

MCQMediumJEE 2024Separation of Variables

The solution curve $y = y(x)$ of the differential equation $(1 + y^2)(1 + \log(x)) \, dx + x \, dy = 0$ , $x > 0$ , passes through the point $(1, 1)$ . If $y(e) = \frac{\alpha - \tan\left(\frac{3}{2}\right)}{\beta + \tan\left(\frac{3}{2}\right)}$ , then $\alpha + 2\beta$ is:

A
$2$
B
$3$
C
$1$
D
$4$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $(1 + y^2)(1 + \log_e x) \, dx + x \, dy = 0$ with $x > 0$ , and the curve passes through $(1,1)$ .

Find: $\alpha + 2\beta$ if

y(e) = \frac{\alpha - \tan\left(\frac{3}{2}\right)}{\beta + \tan\left(\frac{3}{2}\right)}

Rewrite the differential equation in separable form:

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

So,

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

Integrate both sides:

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

Hence,

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

which is equivalently written as

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

Use the condition $(x,y)=(1,1)$ :

\ln 1 + \frac{(\ln 1)^2}{2} + \tan^{-1}(1) = C

0 + 0 + \frac{\pi}{4} = C

So,

\ln x + \frac{(\ln x)^2}{2} + \tan^{-1}y = \frac{\pi}{4}

Now put $x=e$ :

\ln e + \frac{(\ln e)^2}{2} + \tan^{-1}y(e) = \frac{\pi}{4}

1 + \frac{1}{2} + \tan^{-1}y(e) = \frac{\pi}{4}

Therefore,

\tan^{-1}y(e) = \frac{\pi}{4} - \frac{3}{2}

y(e) = \tan\left(\frac{\pi}{4} - \frac{3}{2}\right)

Using

\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

with $A=\frac{\pi}{4}$ and $B=\frac{3}{2}$ ,

y(e) = \frac{1 - \tan\left(\frac{3}{2}\right)}{1 + \tan\left(\frac{3}{2}\right)}

Comparing with

\frac{\alpha - \tan\left(\frac{3}{2}\right)}{\beta + \tan\left(\frac{3}{2}\right)}

we get $\alpha=1$ and $\beta=1$ .

Thus,

\alpha + 2\beta = 1 + 2(1) = 3

Therefore, the correct option is B.

Using substitution $$v = 1 + \log_e x$$

Take

v = 1 + \log_e x

Then

dv = \frac{dx}{x}

The differential equation becomes

\frac{dy}{1+y^2} = -v \, dv

Integrating,

\int \frac{dy}{1+y^2} = \int -v \, dv

\tan^{-1}y = -\frac{v^2}{2} + C

Substitute back $v = 1 + \log_e x$ :

\tan^{-1}y = -\frac{(1+\log_e x)^2}{2} + C

At $(1,1)$ ,

\tan^{-1}(1) = -\frac{(1+\log_e 1)^2}{2} + C

\frac{\pi}{4} = -\frac{1}{2} + C

So,

C = \frac{\pi}{4} + \frac{1}{2}

Hence,

\tan^{-1}y = -\frac{(1+\log_e x)^2}{2} + \frac{\pi}{4} + \frac{1}{2}

For $x=e$ , we have $1+\log_e e = 2$ . Thus,

\tan^{-1}y(e) = -\frac{4}{2} + \frac{\pi}{4} + \frac{1}{2} = \frac{\pi}{4} - \frac{3}{2}

So again,

y(e) = \frac{1 - \tan\left(\frac{3}{2}\right)}{1 + \tan\left(\frac{3}{2}\right)}

Therefore $\alpha=1$ , $\beta=1$ , and $\alpha+2\beta=3$ .

Common mistakes

Treating the equation as linear in $y$ is incorrect because the factor $(1+y^2)$ makes it directly separable. First rearrange to $\frac{dy}{1+y^2}$ and collect the $x$ -terms on the other side.
While integrating $\frac{1+\ln x}{x}$ , students often miss the term $\frac{(\ln x)^2}{2}$ . Use $\int \frac{\ln x}{x} \, dx = \frac{(\ln x)^2}{2}$ .
A common error is using $\tan^{-1}(1)=1$ . This is wrong; $\tan^{-1}(1)=\frac{\pi}{4}$ . The initial condition must be applied with the correct inverse tangent value.
In the tangent subtraction identity, the denominator is $1+\tan A \tan B$ , not $1-\tan A \tan B$ . Use $\tan\left(\frac{\pi}{4}-\frac{3}{2}\right)=\frac{1-\tan\left(\frac{3}{2}\right)}{1+\tan\left(\frac{3}{2}\right)}$ .

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