Embibe Experts Solutions for Chapter: Differential Equation, Exercise 1: WBJEE 2018

The differential equation of the family of curves $y=e^x(A\cos x+B\sin x)$ where $A,B$ are arbitrary constants is

The differential equation representing the family of curves $y^2=2d(x+\sqrt{d}),$ where $d$ is a parameter, is of

General solution of $(x+y{)}^2\frac{dy}{dx}=a^2, a\ne 0$ is $(C$ is an arbitrary constant$)$

The law of motion of a body moving along a straight line is $x=\frac{1}{2}vt$$. x$ being its distance from a fixed point on the line at time $t$ and $v$ is its velocity there. Then

If $x\sin \left(\frac{y}{x}\right)dy=\left[y\sin \left(\frac{y}{x}\right)-x\right]dx,x>0$ and $y\left(1\right)=\frac{\pi }{2},$ then the value of $\cos \left(\frac{y}{x}\right)$ is

The general solution of the differential equation $\left(1+e^{\frac{x}{y}}\right)dx+\left(1-\frac{x}{y}\right)e^{x/y}dy=0$ is
$(C$ is an arbitrary constant$)$

Let $f$ be a differentiable function with $\underset{x\to \infty }{\lim }f\left(x\right)=0.$ If $y^{\text{'}}+y{f}^{\text{'}}\left(x\right)-f\left(x\right)f^{\text{'}}\left(x\right)=0$, $\underset{x\to \infty }{\lim }y\left(x\right)=0,$ then (where $\left.y^{\text{'}}\equiv \frac{dy}{dx}\right)$

Let $y\left(x\right)$ be a solution of $\left(1+x^2\right)\frac{dy}{dx}+2xy-4x^2=0$ and $y\left(0\right)=-1.$ Then $y\left(1\right)$ is equal to