Differential Equations · Mathematics · VITEEE

If $$m$$ and $$n$$ are order and degree of the question $$\left(\frac{d^2 y}{d x^2}\right)^4+8 \frac{\left(d^2 y / d x^2\right)^3}{\left(d^4 y / d x^4\right)^5}+\left(\frac{d^4 y}{d x^4}\right)=x^2+4$$, then $$m-n$$ is equal to

The solution of $$d y / d x=1+x+y+x y$$ is

If $$x d y / d x=x^2+y-2, y(1)=1$$, then $$y(2)$$ is equal to

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

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

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

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

If $$p$$ and $$q$$ are order and degree of the question $$\left(\frac{d^2 y}{d x^2}\right)^4+4 \frac{\left(\frac{d^2 y}{d x^2}\right)^2}{\left(\frac{d^3 y}{d x^3}\right)^3}+\frac{d^3 y}{d x^3}=x^2-1$$, then

The solution of $$\frac{d y}{d x}=1+x+y+x y$$ is

The general solution of the linear differential equation $$\frac{d y}{d x}+\sec x \cdot y=\tan x\left(0 \leq x \leq \frac{\pi}{2}\right)$$ is

On solving the differential equation $$x^2 y d x-\left(x^3+y^3\right) d y=0$$, the value of $$\log y$$ is

The particular solution of the differential equation $$\frac{d y}{d x}+y \cot x=2 x+x^2 \cot x$$, such that $$y(\pi / 2)=0$$ is