Differential Eq...

lf $$f (x)$$ and $$g (x)$$ are two solutions of the differential equation $$a\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{x}^{2}\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}$$, then $$f (x) - g (x)$$ is the solution of

Solution of $$\dfrac{d^{2}y}{dx^{2}}$$= $$\log x$$ is:

lf the solution of the differential equation
$$\displaystyle \frac{1}{\sin^{-1}x}(\frac{dy}{dx})=1$$ is $$y=x \sin^{-1}x+f(x)+c$$ then f (x) is

The solution of $$\displaystyle \frac{dy}{dx}+\frac{x(1+y^{3})}{y^{2}(1+x^{2})}=0$$ is:

The solution of $$(1-x^{2})\displaystyle \frac{dy}{dx}+xy=xy^{2}$$ is:

Solution of $$\>y-x\displaystyle \frac{dy}{dx}=5\left(y^{2}+\frac{dy}{dx}\right)$$, is:

If $$ \dfrac{dy}{dx}=xy+2x+3y+6$$, then find the value of $$y(-1)-e^2y(-3)$$.

A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k, then show that the differential equation describing such curves Is, $$\displaystyle y\frac{dy}{dx}=\pm \sqrt{k^{2}-y^{2}}$$. Find the equation of such a curve passing through (0.k), the curve is a: 

The solution of differential equation $$(e^x + 1) y dy = (y + 1) e^x dx$$ is :

The solution of differential equation  $$(e^x + 1)y dy = (y + 1) e^x  dx$$ is: