The solution curve of the differential equation: y dxdy = x [( (x)) - ( (y)) + 1 ], with x 0, y 0, passing through the point (e, 1)

Given: y dxdy = x [( x) - ( y) + 1 ] with x 0, y 0 and the curve passes through (e,1). Find: The relation satisfied by the solution curve. Rewrite the equation in terms of dxdy: 1x dxdy = x - y + 1y Now observe that ddy( x) = 1x dxdy So the differential equation becomes ddy( x) = x - y + 1y Let z = x Then the equation becomes dzdy = z - y + 1y Set u = z - y Then d dy = dzdy - 1y = z - y + 1y - 1y = y Hence d = dyy Integrating, | | = y + C So | | = Cy that is, | x - y| = Cy Using logarithm property, | ( xy) | = Cy Now apply the point (e,1): | ( e1) | = C(1) Thus 1 = C Therefore | x - y| = y which is equivalent to | ( xy) | = y Since ( xy) = x - y this matches the option written as | xy | = y in the source only in intent, but the algebraically correct relation is | (x/y)| = y. Based on the provided options and answer key, the most defensible choice is C. Therefore, the correct option is C.

The solution curve of the differential equation: y dxdy = x [( (x… — Solution | JEE Mathematics

MCQMediumJEE 2024Separation of Variables

The solution curve of the differential equation: $y \frac{dx}{dy} = x \left[(\ln(x)) - (\ln(y)) + 1\right]$ , with $x > 0$ , $y > 0$ , passing through the point $(e, 1)$

A
$\left|\frac{\ln(y)}{x}\right| = x$
B
$\left|\frac{\ln(y)}{x}\right| = y^2$
C
$\left|\frac{\ln(x)}{y}\right| = y$
D
$2\left|\frac{\ln(x)}{y}\right| = y + 1$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: $y \frac{dx}{dy} = x\left[(\ln x) - (\ln y) + 1\right]$ with $x > 0$ , $y > 0$ and the curve passes through $(e,1)$ .

Find: The relation satisfied by the solution curve.

Rewrite the equation in terms of $\frac{dx}{dy}$ :

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

Now observe that

\frac{d}{dy}(\ln x) = \frac{1}{x}\frac{dx}{dy}

So the differential equation becomes

\frac{d}{dy}(\ln x) = \frac{\ln x - \ln y + 1}{y}

Using a substitution

Let

z = \ln x

Then the equation becomes

\frac{dz}{dy} = \frac{z - \ln y + 1}{y}

Set

u = z - \ln y

Then

\frac{d\nu}{dy} = \frac{dz}{dy} - \frac{1}{y} = \frac{z - \ln y + 1}{y} - \frac{1}{y} = \frac{\nu}{y}

Hence

\frac{d\nu}{\nu} = \frac{dy}{y}

Integrating,

\ln|\nu| = \ln y + C

|\nu| = Cy

that is,

|\ln x - \ln y| = Cy

Using logarithm property,

\left|\ln\left(\frac{x}{y}\right)\right| = Cy

Now apply the point $(e,1)$ :

\left|\ln\left(\frac{e}{1}\right)\right| = C(1)

Thus

1 = C

Therefore

|\ln x - \ln y| = y

which is equivalent to

\left|\ln\left(\frac{x}{y}\right)\right| = y

Since

\ln\left(\frac{x}{y}\right) = \ln x - \ln y

this matches the option written as $\left|\frac{\ln x}{y}\right| = y$ in the source only in intent, but the algebraically correct relation is $|\ln(x/y)| = y$ . Based on the provided options and answer key, the most defensible choice is C.

Therefore, the correct option is C.

Common mistakes

Treating $\ln x - \ln y$ as $\frac{\ln x}{y}$ is incorrect. Logarithms do not distribute that way. Use $\ln x - \ln y = \ln\left(\frac{x}{y}\right)$ instead.
Forgetting to convert $\frac{1}{x}\frac{dx}{dy}$ into $\frac{d}{dy}(\ln x)$ makes the equation look harder than it is. Recognize this derivative form before attempting separation.
Not using the initial point $(e,1)$ to determine the constant leaves only a family of curves. After integration, always substitute the given point to find the specific solution curve.

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Answer

Step-by-step solution

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