What is the solution of \(y\,dx-x\,dy-y^{2}\cos x\,dx=0\)?
\(y=\dfrac{x}{c+\sin x}\)
Dividing the equation by \(y^{2}\) gives \(\dfrac{y\,dx-x\,dy}{y^{2}}=\cos x\,dx\), and the left side is exactly \(d\!\left(\dfrac{x}{y}\right)\). Integrating both sides gives \(\dfrac{x}{y}=\sin x+c\), so \(y=\dfrac{x}{c+\sin x}\).
What is the general solution of the differential equation ydx – (x + 2y 2) dy = 0?
If xdy = y(dx + ydy) ; y(1) = 1 and y(x) > 0, then what is y(-3) equal to?
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:
What is the solution of the differential equation (dy − dx) + cos x(dy + dx) = 0 ?
What is the solution of the following differential equation?
\(\rm \ln\left(\frac{dy}{dx}\right)+y = x\)
The solution of the differential equation \(\frac{{dy}}{{dx}} = \cos \left( {y - x} \right) + 1\) is
What is the solution of the differential equation
\(\frac{{dx}}{{dy}} + \frac{x}{y} - {y^2} = 0\) ?
Where c is an arbitrary constant.
The solution of the differential equation dy = (1 + y 2) dx is
What is the general solution of the differential equation \(\frac{{dy}}{{dx}} + \frac{x}{y} = 0\) ?
The equation of motion of a one-dimensional forced harmonic oscillator in the presence of a dissipative force is described by \(\frac{{{{\rm{d}}^{\rm{2}}}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^{\rm{2}}}}}\,{\rm{ + }}\,{\rm{10}}\frac{{{\rm{dx}}}}{{{\rm{dt}}}}\,{\rm{ + }}\,{\rm{16x}}\,{\rm{ = }}\,{\rm{6t}}{{\rm{e}}^{{\rm{ - 8t}}}}{\rm{ + }}\,{\rm{4}}{{\rm{t}}^{\rm{2}}}{{\rm{e}}^{{\rm{ - 2t}}}}\) The general form of the particular solution, in terms of constants A, B etc., is
What is the general solution of the differential equation ydx – (x + 2y 2) dy = 0?
If xdy = y(dx + ydy) ; y(1) = 1 and y(x) > 0, then what is y(-3) equal to?
If y(x) is a solution of the differential equation \(\frac{{dy}}{{dx}} + 4xy = {x^3},y(0) = 0\) then \(\mathop {\lim }\limits_{x \to 0} y(x)\) is