If the slope of the tangent to the curve at any point P(x, y) is (y)/(x)- ^ 2 (y)/(x) , then the equation of a curve passing through (1, (π)/(4)) is

Correct answer: A. ((y)/(x))+ x=1 According to the condition, dy dx =(y)/(x)- ^ 2 (y)/(x) (i) This is a homogeneous differential equation Substituting y=v x , we get array l v+x dv dx =v- ^ 2 v x dv dx =- ^ 2 v ∫ ^ 2 vdv=-∫ dx x v=- x+C (y)/(x)+ x=C array Substituting x=1, y=(π)/(4) , we get C=1 Thus, we get ((y)/(x))+ x=1 which is the required solution.

BITSAT 2021

Exam: BITSAT
Year: 2021
Subject: Mathematics
Chapter: Differential Equations

BITSAT 2021 · Mathematics · Differential Equations

If the slope of the tangent to the curve at any point P(x, y) is...

If the slope of the tangent to the curve at any point

\mathrm{P}(x, y)

\frac{y}{x}-\cos ^{2} \frac{y}{x}

, then the equation of a curve passing through

\left(1, \frac{\pi}{4}\right)

Options

A $\tan \left(\frac{y}{x}\right)+\log x=1$
B $\tan \left(\frac{y}{x}\right)+\log y=1$
C $\tan \left(\frac{x}{y}\right)+\log x=1$
D $\tan \left(\frac{x}{y}\right)+\log y=1$

Correct answer

A. $\tan \left(\frac{y}{x}\right)+\log x=1$

Step-by-step solution

According to the condition,

\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y}{x}-\cos ^{2} \frac{y}{x} \ldots \text{ (i) } This is a homogeneous differential equation Substituting

y=v x

, we get

\begin{array}{l} v+x \frac{\mathrm{dv}}{\mathrm{dx}}=v-\cos ^{2} v \\ \Rightarrow x \frac{\mathrm{dv}}{\mathrm{dx}}=-\cos ^{2} v \\ \Rightarrow \int \sec ^{2} \mathrm{vdv}=-\int \frac{\mathrm{dx}}{\mathrm{x}} \\ \Rightarrow \tan v=-\log x+\mathrm{C} \\ \Rightarrow \tan \frac{y}{x}+\log x=\mathrm{C} \end{array}

Substituting

\mathrm{x}=1, y=\frac{\pi}{4}

, we get

\mathrm{C}=1

Thus, we get

\tan \left(\frac{y}{x}\right)+\log x=1$ which is the required solution.

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