Question

Assertion :The solution of $\frac{x+\frac{dy}{dx}}{y-x\frac{dy}{dx}}=\frac{x{\cos }^2(x^2+y^2)}{y^3}$ is $\tan (x^2+y^2)=\frac{x^2}{y^2}+c$ Reason: $xdx+ydy=\frac{1}{2}d(x^2+y^2)$ & $\frac{ydx-xdy}{y^2}=d\left(\frac{x}{y}\right)$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Given $\frac{x+\frac{dy}{dx}y}{y-x\frac{dy}{dx}}=\frac{x{\cos }^{2}x(x^2+y^2)}{y^3}$

$\Rightarrow$ $\frac{xdx+ydy}{\left(\frac{ydx-xdy}{y^3}\right)y^2}=\frac{x}{y^3}{\cos }^2(x^2+y^2)$

$\Rightarrow$ ${\sec }^{2}x(x^2+y^2)\frac{1}{2}d(x^2+y^2)=\frac{x}{y}d\left(\frac{x}{y}\right)$

Integrating both sides, we get

$\Rightarrow$ $\frac{1}{2}\tan (x^2+y^2)=\frac{1}{2}{\left(\frac{x}{y}\right)}^2+c$

$\Rightarrow$ $\tan (x^2+y^2)={\left(\frac{x}{y}\right)}^2+k$