G Tewani Solutions for Chapter: Differential Equations, Exercise 2: DPP

The general solution of $x\left(\frac{dy}{dx}\right)+\left({\log }_{e}x\right)y=x^{1-\frac{1}{2}\log x}$ is:

The general solution of the differential equation $(1+\tan y)(dx-dy)+2xdy=0$ is

Solution of differential equation $x^{2}y-x^3\frac{dy}{dx}=y^4\cos x$ is:

Suppose that a solution of the differential equation $\left(x{y}^3+x^2{y}^7\right)\frac{dy}{dx}=1$ satisfies the initial condition $y\left(\frac{1}{4}\right)=1.$ Then the value of $\frac{dy}{dx}$ when $y=-1$ is 

The general solution of the differential equation $\frac{dy}{dx}=y\tan x-y^2\sec x$ is

Let $\frac{dy}{dx}+y=f\left(x\right)$ where $y$ is a continuous function of $x$ with $y\left(0\right)=1$ and

 $f\left(x\right)=\left\{\begin{array}{ll}{e}^{-x}& \text{ if }0\le x\le 2\\ e^{-2}& \text{if }x>2\end{array}\right.$.

Which of the following then hold(s) good?

A differentiable function satisfies $f\left(x\right)={\int }_0^x(f\left(t\right)\cos t-\cos \left(t-x\right))dt$. Which of the following hold(s) good?

$y=f\left(x\right)$ has differential equation $y\left(2xy+e^{\mathrm{x}}\right)dx-e^{\mathit{x}}dy=0$, which is passing through the point $(0, 1).$ Then which of the following is/are true about the function?