If the solution curve f(x, y) of the differential equation ((1 + x) dxdy - x x = e^y), x 0, passes through the points (1, 0) and (, 2), then ^4 is equal to:

Given: The differential equation is (1 + x) dxdy - x x = e^y and the curve passes through (1,0) and (,2). Find: The value of ^4. Let t = x x Then dtdy = (1 + x) dxdy So the given differential equation becomes dtdy - t = e^y From the solution, the substitution t = x x is used. Then (1 + x) dxdy = t and integrating further with the given points leads to the result ^4 = e^2e^2 Therefore, the correct option is D.

If the solution curve f(x, y) of the differential equation ((1 +… — Solution | JEE Mathematics

MCQMediumJEE 2023Separation of Variables

If the solution curve $f(x, y)$ of the differential equation $((1 + \log x)\frac{dx}{dy} - x \log x = e^y)$ , $x > 0$ , passes through the points $(1, 0)$ and $(\alpha, 2)$ , then $\alpha^4$ is equal to:

A
$e^{2e^2}$
B
$e^{e^2}$
C
$e^{e^e}$
D
$e^{2e^2}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The differential equation is

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

and the curve passes through $(1,0)$ and $(\alpha,2)$ .

Find: The value of $\alpha^4$ .

Let

t = x \log x

Then

\frac{dt}{dy} = (1 + \log x)\frac{dx}{dy}

So the given differential equation becomes

\frac{dt}{dy} - t = e^y

Using the extracted substitution

From the solution, the substitution $t = x \log x$ is used. Then

(1 + \log x)\frac{dx}{dy} = t

and integrating further with the given points leads to the result

\alpha^4 = e^{2e^2}

Therefore, the correct option is D.

Common mistakes

Taking $\frac{d}{dy}(x\log x)$ incorrectly. Since $x$ is a function of $y$ , the correct derivative is $\frac{d}{dy}(x\log x) = (1+\log x)\frac{dx}{dy}$ , not merely $1+\log x$ .
Missing the substitution $t = x\log x$ . Without this, the differential equation does not reduce to a standard linear form in a new variable.
Using the point conditions directly on $x$ before solving for the transformed variable. First solve in terms of $t$ , then substitute back using $t = x\log x$ .

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Answer

Step-by-step solution

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