List of top Mathematics Questions on Differential Equations asked in TS PGECET

The initial value problem \[ (x-x^2)\frac{dy}{dx} = (2x-1)y, \qquad y(x_0)=y_0, \] has a unique solution if \((x_0,y_0)\) equals to

The particular integral of the differential equation \[ (D^2+1)y = \sin x\,\sin2x \] is

The general solution of \[ \log\left(\frac{dy}{dx}\right)=ax+by \] is

If \(y_1,\;y_2\) are two dependent solutions of a second order linear homogeneous differential equation then \(y_1y_2'-y_2y_1'\) is

The solution of \[ \frac{dy}{dx} + \left[2(x-1)\sin x+x(x-2)\cos x\right]y = \frac{e^{2x\sin x}}{e^{x^2\sin x}}, \] at \(x=2\) if \(y(\pi)=0\) is

The initial value problem \[ (x-x^2)\frac{dy}{dx} = (2x-1)y, \qquad y(x_0)=y_0, \] has a unique solution if \((x_0,y_0)\) equals to

The particular integral of the differential equation \[ (D^2+1)y = \sin x\,\sin2x \] is

The general solution of \[ \log\left(\frac{dy}{dx}\right)=ax+by \] is

If \(y_1,\;y_2\) are two dependent solutions of a second order linear homogeneous differential equation then \(y_1y_2'-y_2y_1'\) is

The solution of \[ \frac{dy}{dx} + \left[2(x-1)\sin x+x(x-2)\cos x\right]y = \frac{e^{2x\sin x}}{e^{x^2\sin x}}, \] at \(x=2\) if \(y(\pi)=0\) is

If the integrating factor of $\frac{dy}{dx} + \left[(x^2 - 2x)\cos x + 2(x - 1)\sin x\right]y = x^2$ is $e^{f(x)}$, then $f(3) =$}

If the particular integral of \( y'' - 4y' = x^2 e^{2x} \) is in the form \( y_p = e^{2x} y(x) \), then \( y(x) \) is

If the integrating factor of $\frac{dy}{dx} + \left[(x^2 - 2x)\cos x + 2(x - 1)\sin x\right]y = x^2$ is $e^{f(x)}$, then $f(3) =$}

If the particular integral of \( y'' - 4y' = x^2 e^{2x} \) is in the form \( y_p = e^{2x} y(x) \), then \( y(x) \) is

The particular integral of $\left(D^4 - D^3 - 9D^2 - 11D - 4\right)y = e^{-x}$, where $D = \frac{d}{dx}$, is:}

Evaluate \[ \int_{-1}^{1}\frac{x^{2}\sin x}{x^{4}+1}\,dx. \]

The maximum value of \(x^{4}y^{3}\) such that \(x+y=42\) exists at \(x=\alpha,\; y=\beta\). Then \(\dfrac{\alpha}{\beta}\) in its lowest form is

If the solution of \[ \frac{d^2y}{dt^2} = -\frac54y+\frac{dy}{dt}, \] satisfying \[ y(0)=1, \qquad \left(\frac{dy}{dt}\right)_{t=0}=0, \] is \(y(t)\), then \(y(\pi)=\)