If k is constant, the general solution of \(\frac{dy}{dx} - \frac{y}{x}=1\) will be in the form of

If k is constant, the general solution of \(\frac{dy}{dx} - \frac{y}{

1.	If k is constant, the general solution of \(\frac{dy}{dx} - \frac{y}{x}=1\) will be in the form of
A.	y = x ln (kx)
B.	y = k ln (kx)
C.	y = xk ln (k)
D.	y = x ln (x)
Answer» B. y = k ln (kx)

For \(\frac{{{d^2}y}}{{d{x^2}}} + 4\frac{{dy}}{{dx}} + 3y = 3{e^{2x}}\), the particular integral is :
Consider the differential equation given below:\(\dfrac{dy}{dx} + \dfrac{x}{1-x^2} y= x\sqrt{y}\)The integrating factor of the differential equation is:
General solution of the equation (x3 + 3xy2)dx + (3x2y + y3)dy = 0 is (c is a constant)
For the equation \(\frac{{dy}}{{dx}} + 7{x^2}y = 0\), if y(0) = \(\frac{{3}}{{7}}\), then the value of y(1) is
If \(\rm \dfrac{d^2y}{dx^2}+\sin x = 0\), then the solution of this differential equation is -
\(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {c ^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\) represents the equation for
Consider the following partial differential equation for with the constant c > 1:Solution of this equation is \(\frac{{\partial u}}{{\partial y}} + c\frac{{\partial u}}{{\partial x}} = 0\)
Determine the degree and order of the given differential equation respectively. \({\rm{y}} = {\rm{x}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} + {\left( {\frac{{{\rm{dx}}}}{{{\rm{dy}}}}} \right)^2}?\)
If the characteristic equation of the differential equation \(\frac{{{d^2}y}}{{d{x^2}}} + 2\alpha \frac{{dy}}{{dx}} + y = 0\) has two equal roots, then the values of α are
A particle starts from origin with a velocity (in m/s) given by the equation \(\rm \frac{dx}{dt}=x+1\). The time (in seconds) taken by the particle to traverse a distance of 24 m is: