Differential Equations PYQ - Calculus | Mathematics

The solution of the differential equation \(\frac{d^2 y}{d x^2}+y=0\) is

The equation of the curve passing through the point \(\left(0, \frac{\pi}{4}\right)\) and satisfying the differential equation \(\left(e^x \tan y\right) d x\left.+\left(1+e^x\right) \sec ^2 y\right) d y=0\) is given by

If \(y=\sin (\sin x)\) and \(y^{\prime \prime}+f(x) \cdot y^{\prime}+g(x) \cdot y=0\), then \(f(x) \cdot g(x)\) is equal to

The solution of the differential equation \(2x\left(\frac{dy}{dx}\right)-y=4\) represents a family of

By eliminating the arbitrary constants from \(y=(a+b) \sin (x+c)-d e^{x+e+f}\), then differential equation has order of

If \(2 x-y+C \log (|x-2 y-4|)=k\) is the general solution of \(\frac{d y}{d x}=\frac{2 x-4 y-5}{x-2 y+2}\), then \(C=\)

If \(l\) and \(m\) are order and degree of a differential equation of all the straight lines at constant distance of \(P\) units from the origin, then \(l m^2+l^2 m=\)

If the solution of \(\frac{d y}{d x}-y \log _e 0.5=0, y(0)=1\), and \(y(x) \rightarrow k\), as \(x \rightarrow \infty\), then \(k=\)

\(y=A e^x+B e^{-2 x}\) satisfies which of the following differential equations?

Assertion (A) Order of the differential equations of a family of circles with constant radius is two.

Reason (R) An algebraic equation having two arbitrary constants is general solution of a second order differential equation.

The general solution of the differential equation \(\frac{d y}{d x}=\cos ^2(3 x+y)\) is \(\tan ^{-1}\left(\frac{\sqrt{3}}{2} \tan (3 x+y)\right)=f(x)\). Then, \(f(x)=\)

If the general solution of the differential equation \(\cos ^2 x \frac{d y}{d x}+y=\tan x\) is \(y=\tan x-1+C e^{-\tan x}\) satisfies \(y\left(\frac{\pi}{4}\right)=1\), then \(C=\)