Differential Equations
Assume that a spherical rain drop evaporates at a rate proportional to its surface area. If k is a positive constant then differential equation involving the rate of change of the radius of the rain drop is
Find the differential equation of all the parabola having axis parallel to the $$x-$$axis.
Equation of parabola having axis parallel to x-axis is given by,
$$(y-\beta )^2=a(x-\alpha )$$
differentiating on both sides, we get,
$$2(y-\beta )=\dfrac{dy}{dx}=a$$
again differentiating on both sides, we get,
$$2(y-\beta )=\dfrac{d^2y}{dx^2}+2\left ( \dfrac{dy}{dx} \right )^2=0$$
eliminating $$\beta $$ from the above equations, we get,
$$a\dfrac{d^2y}{dx^2}+2\left ( \dfrac{dy}{dx} \right )^3=0$$
Is the required differential equation.
Assume that a spherical rain drop evaporates at a rate proportional to its surface area. If k is a positive constant then differential equation involving the rate of change of the radius of the rain drop is
The differential equation of all parabolas whose axes are parallel to $$\mathrm{y}$$ -axis is:
$$\displaystyle \left ( x^{3}-y^{3} \right )dx+xy^{2}dy= 0.$$ Solving this we get $$\displaystyle \frac{k}{x}=e^{y^{m}/nx^{r}} $$.Find $$m+n+r$$ ?
If $$y=e^{m sin^{-1}x}$$, then show that $$(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}-m^2y=0$$.
If $$\displaystyle 2x=y^{\tfrac{1}{5}}+y^{-\tfrac{1}{5}}$$ and $$\displaystyle (x^2-1)\dfrac{d^2y}{dx^2}+\lambda x\dfrac{dy}{dx}+ky=0$$, then $$\lambda +k$$ is equal to:
Let $$(x,y,z)$$ be points with integer coordinates satisfying the system of homogeneous equations: $$3x-y-z=0$$ $$-3x+z=0$$ $$-3x+2y+z=0$$. Then the number of such points which lie inside a sphere of radius $$10$$ centered at the origin is
The differential equation of the family of circles touching y-axis at the origin is:
The differential equation of all parabolas whose axis is y-axis is:
If $$ y={ \left( \tan ^{ -1 }{ x } \right) }^{ 2 }$$, then $${ \left( { x }^{ 2 }+1 \right) }^{ 2 }\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +2x\left( { x }^{ 2 }+1 \right) \cfrac { dy }{ dx } =$$
if $$y = A{e^{ - kt}}\cos (pt + c)$$ , then prove that $${{{d^2}y} \over {d{t^2}}} + 2k{{dy} \over {dt}} + {n^2}y = 0$$ , where $${n^2} = {p^2} + {k^2}$$