A curve passing through $$\left( {2,\,3} \right)$$ and satisfying the differential equation $$\int_0^x {ty\left( t \right)dt = {x^2}y\left( x \right),\,\left( {x > 0} \right)} {\text{ is}}\,{\text{:}}$$
A.
$${x^2} + {y^2} = 13$$
B.
$${y^2} = \frac{9}{2}x$$
C.
$$\frac{{{x^2}}}{8} + \frac{{{y^2}}}{{18}} = 1$$
D.
$$xy = c$$
Answer :
$$xy = c$$
Solution :
$$\eqalign{
& {\text{Differentiate }}xy\left( x \right) = {x^2}y'\left( x \right) + 2xy\left( x \right) \cr
& {\text{or }}xy\left( x \right) + {x^2}y'\left( x \right) = 0 \cr
& {\text{or }}x\frac{{dy}}{{dx}} + y = 0 \cr
& {\text{or }}\ln \,y + \ln \,x = \ln \,c \cr
& {\text{or }}xy = c \cr} $$
Releted MCQ Question on Calculus >> Differential Equations
Releted Question 1
A solution of the differential equation $${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$$ is-
If $$y\left( t \right)$$ is a solution $$\left( {1 + t} \right)\frac{{dy}}{{dt}} - ty = 1$$ and $$y\left( 0 \right) = - 1,$$ then $$y\left( 1 \right)$$ is equal to-