Homogeneous equations

Let $x=x(y)$ be the solution of the differential equation $2y^2\dfrac{dx}{dy}-2xy+x^2=0$, $y>1$, $x(e)=e$. Then $x\left(e^2\right)$ is equal to :

Let $y=y(x)$ be the solution of the differential equation $x\sin\left(\dfrac yx\right)dy=\left(y\sin\left(\dfrac yx\right)-x\right)dx$, $y(1)=\dfrac\pi2$ and let $\alpha=\cos\left(\dfrac{y\left(e^{12}\right)}{e^{12}}\right)$. Then the number of integral value of $p$, for which the equation $x^2+y^2-2px+2py+\alpha+2=0$ represents a circle of radius $r\le6$, is ________.

Let $y(x)$ be the solution of the differential equation

$$x^2\frac{dy}{dx} + xy = x^2 + y^2,\qquad x > \frac1e,$$

satisfying $y(1) = 0$. Then the value of $2\dfrac{(y(e))^2}{y(e^2)}$ is _____.

If the solution curve $y=y(x)$ of the differential equation $y^2dx+\left(x^2-xy+y^2\right)dy=0$, which passes through the point $(1,1)$ and intersects the line $y=\sqrt3x$ at the point $\left(\alpha,\sqrt3\alpha\right)$, then value of $\log_e\left(\sqrt3\alpha\right)$ is equal to :

If $y=y(x)$ is the solution of the differential equation $2x^2\dfrac{dy}{dx}-2xy+3y^2=0$ such that $y(e)=\dfrac e3$, then $y(1)$ is equal to

Let the solution curve $y=y(x)$ of the differential equation
$$\left[\frac x{\sqrt{x^2-y^2}}+e^{\frac yx}\right]x\frac{dy}{dx}=x+\left[\frac x{\sqrt{x^2-y^2}}+e^{\frac yx}\right]y$$
pass through the points $(1,0)$ and $(2\alpha,\alpha)$, $\alpha>0$. Then $\alpha$ is equal to

Let $x=x(y)$ be the solution of the differential equation $2y\,e^{x/y^2}dx+\left(y^2-4xe^{x/y^2}\right)dy=0$ such that $x(1)=0$. Then, $x(e)$ is equal to :

Let the solution curve of the differential equation $x\dfrac{dy}{dx}-y=\sqrt{y^2+16x^2}$, $y(1)=3$ be $y=y(x)$. Then $y(2)$ is equal to :

The general solution of the differential equation $\left(x-y^2\right)dx+y\left(5x+y^2\right)dy=0$ is :

Let $y=y(x)$ be the solution of the differential equation
$$\frac{dy}{dx}=\frac{4y^3+2yx^2}{3xy^2+x^3},\quad y(1)=1.$$
If for some $n\in\mathbb N$, $y(2)\in[n-1,n)$, then $n$ is equal to ______.

Let the solution curve of the differential equation $x\,dy=\left(\sqrt{x^2+y^2}+y\right)dx$, $x>0$, intersect the line $x=1$ at $y=0$ and the line $x=2$ at $y=\alpha$. Then the value of $\alpha$ is :

If the solution curve of the differential equation $\dfrac{dy}{dx}=\dfrac{x+y-2}{x-y}$ passes through the points (2, 1) and (k + 1, 2), k > 0, then

If $y=y(x)$ satisfies the differential equation $x\,dy+\left(x\tan\left(\dfrac yx\right)-y\right)dx=0$, x > 0, and $y(1)=\dfrac\pi2$, then a value of $y(\sqrt2)$ is

A solution curve of the differential equation $(x^2+xy+4x+2y+4)\dfrac{dy}{dx}-y^2=0$, $x>0$, passes through the point $(1,3)$. Then the solution curve

If the general solution of the differential equation $y'=\dfrac yx+\Phi\left(\dfrac xy\right)$, for some function $\Phi$, is given by $y\ln|cx|=x$, where $c$ is an arbitrary constant, then $\Phi(2)$ is equal to :

A curve passes through the point $\left(1,\frac\pi6\right)$. Let the slope of the curve at each point $(x,y)$ be $\dfrac yx+\sec\left(\dfrac yx\right)$, $x>0$. Then the equation of the curve is