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

If y = y(x) is the solution of the differential equation \left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0 and y (0) = 0, then $$6\left(y^{\prime }(0)+{\left(y\left({\log }_e\sqrt{3}\right)\right)}^2\right)$$ is equal to

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

Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 $$\tan x(\cos x-y)$$. If the curve passes through the point \left( {{\pi \over 4},0} \right), then the value of $$\underset{0}{\overset{\pi /2}{\int }}ydx$$ is equal to :

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

Let y = y(x) be the solution of the differential equation x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0, $$x>1$$, with $$y(2)=-2$$. Then y(3) is equal to :

If the solution curve of the differential equation $$(({\tan }^{-1}y)-x)dy=(1+y^2)dx$$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

Let {{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :

If {{dy} \over {dx}} + {{{2^{x - y}}({2^y} - 1)} \over {{2^x} - 1}} = 0, x, y > 0, y(1) = 1, then y(2) is equal to :

If $$y=y(x)$$ is the solution of the differential equation x{{dy} \over {dx}} + 2y = x\,{e^x}, $$y(1)=0$$ then the local maximum value of the function $$z(x)=x^{2}y(x)-e^x,x\in R$$ is :

If the solution of the differential equation {{dy} \over {dx}} + {e^x}\left( {{x^2} - 2} \right)y = \left( {{x^2} - 2x} \right)\left( {{x^2} - 2} \right){e^{2x}} satisfies $$y(0)=0$$, then the value of y(2) is _______________.

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

Let $$g:(0,\infty )\to R$$ be a differentiable function such that \int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c}, for all x > 0, where c is an arbitrary constant. Then :

Let $$y=y(x)$$ be the solution of the differential equation $$(x+1)y^{\prime }-y=e^{3x}(x+1{)}^2$$, with y(0) = {1 \over 3}. Then, the point x = - {4 \over 3} for the curve $$y=y(x)$$ is :

If the solution curve $$y=y(x)$$ of the differential equation $$y^{2}dx+(x^2-xy+y^2)dy=0$$, which passes through the point (1, 1) and intersects the line $$y=\sqrt{3}x$$ at the point $$(\alpha ,\sqrt{3}\alpha )$$, then value of $${\log }_e(\sqrt{3}\alpha )$$ is equal to :

If x = x(y) is the solution of the differential equation y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0; then x(e) is equal to :

Let {{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R, represents a circle with center ($$\alpha$$, $$\beta$$). Then, $$\alpha$$ + 2$$\beta$$ is equal to :

The slope of the tangent to a curve $$C:y=y(x)$$ at any point $$(x,y)$$ on it is $$\frac{2{\mathrm{e}}^{2x}-6{\mathrm{e}}^{-x}+9}{2+9{\mathrm{e}}^{-2x}}$$. If $$C$$ passes through the points $$\left(0,\frac{1}{2}+\frac{\pi }{2\sqrt{2}}\right)$$ and $$\left(\alpha ,\frac{1}{2}{\mathrm{e}}^{2\alpha }\right)$$, then $${\mathrm{e}}^{\alpha }$$ is equal to :

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

Let a smooth curve $$y=f(x)$$ be such that the slope of the tangent at any point $$(x,y)$$ on it is directly proportional to $$\left(\frac{-y}{x}\right)$$. If the curve passes through the points $$(1,2)$$ and $$(8,1)$$, then $$\left|y\left(\frac{1}{8}\right)\right|$$ is equal to