Let the solution curve of the differential eq | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the differential equation $x \frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$.Divide by $x$: $\frac{dy}{dx} - \frac{y}{x} = \sqrt{(\frac{y}{x})^2 + 1

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) Given the differential equation $x \frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$.Divide by $x$: $\frac{dy}{dx} - \frac{y}{x} = \sqrt{(\frac{y}{x})^2 + 16}$.Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting this into the equation: $v + x \frac{dv}{dx} - v = \sqrt{v^2 + 16}$.$x \frac{dv}{dx} = \sqrt{v^2 + 16} \Rightarrow \int \frac{dv}{\sqrt{v^2 + 16}} = \int \frac{dx}{x}$.Integrating both sides: $\ln|v + \sqrt{v^2 + 16}| = \ln|x| + \ln|C|$.$v + \sqrt{v^2 + 16} = Cx \Rightarrow \frac{y}{x} + \sqrt{\frac{y^2}{x^2} + 16} = Cx$.$y + \sqrt{y^2 + 16x^2} = Cx^2$.Using $y(1) = 3$: $3 + \sqrt{3^2 + 16(1)^2} = C(1)^2 \Rightarrow 3 + \sqrt{25} = C \Rightarrow C = 8$.Thus,$y + \sqrt{y^2 + 16x^2} = 8x^2$.For $x = 2$: $y + \sqrt{y^2 + 16(4)} = 8(4) = 32$.$y + \sqrt{y^2 + 64} = 32 \Rightarrow \sqrt{y^2 + 64} = 32 - y$.Squaring both sides: $y^2 + 64 = 1024 - 64y + y^2$.$64y = 960 \Rightarrow y = 15$.

Let the solution curve of the differential equation $x \frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$ with the initial condition $y(1) = 3$ be $y = y(x)$. Then the value of $y(2)$ is equal to:

Similar Questions

One card is drawn at random from a well-shuffled deck of $52$ cards. In which of the following cases are the events $E$ and $F$ independent?
$E:$ 'The card drawn is a king or a queen'
$F:$ 'The card drawn is a queen or a jack'

The general solution of the differential equation $(x+y) y dx + (y-x) x dy = 0$ is

Show that the differential equation $\left(x^{2}+x y\right) d y=\left(x^{2}+y^{2}\right) d x$ is a homogeneous equation and find its solution.

The particular solution of the differential equation,$x y \frac{dy}{dx} = x^2 + 2y^2$ when $y(1) = 0$ is

The slope of the tangent at $(x, y)$ to the curve passing through $(2, 1)$ is $\frac{x^2+y^2}{2xy}$. Find the equation of the curve.

Vedclass Test Series

Exam Paper Generator

Online Exam Module