VITEEE 2023 | Differential Equations Question 4 | Mathematics

The solution of differential equation $$y y^{\prime}=x\left(\frac{y^2}{x^2}+\frac{f\left(y^2 / x^2\right)}{f^{\prime}\left(y^2 / x^2\right)}\right)$$ is

$$x^2 f\left(\frac{y^2}{x^2}\right)=c$$

$$x^2 f\left(\frac{y^2}{x^2}\right)=c^2 y^2$$

$$f\left(\frac{y^2}{x^2}\right)=c x^2$$

$$f\left(\frac{y^2}{x^2}\right)=\frac{c y}{x}$$

VITEEE 2022

MCQ (Single Correct Answer)

-0

Find the solution of equation $$\frac{d y}{d x}=\frac{1}{\cos (x+y)}$$

$$\cot \left(\frac{x+y}{2}\right)+c$$

$$\sec \left(\frac{x+y}{2}\right)+c$$

$$\tan \left(\frac{x+y}{2}\right)+c$$

$$-\cot \left(\frac{x-y}{2}\right)+c$$

VITEEE 2022

MCQ (Single Correct Answer)

-0

Find the solution of $$\frac{d y}{d x}=\frac{1}{\cos (x-y)}$$

$$-\cot \left(\frac{x+y}{2}\right)+c$$

$$\operatorname{cosec}^2\left(\frac{x+y}{2}\right)+c$$

$$\tan \left(\frac{x+y}{2}\right)+c$$

$$\cot \left(\frac{x-y}{2}\right)+c$$

VITEEE 2022

MCQ (Single Correct Answer)

-0

The solution of the equation $$\frac{d y}{d x}+x(x+y)=x^3(x+y)^3-1$$ is

$$\frac{1}{(x+y)^2}=x^2+1+c e^x$$

$$\frac{1}{(x+y)^2}=x^2+1+c e^{x^2}$$

$$\frac{1}{x+y}=x+1+c e^{x^2}$$

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