The solution of the differential equation 3 x | vedclass

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

Question

(A) Given differential equation: $3 x y' - 3 y + \sqrt{x^2 - y^2} = 0$. Dividing by $x$: $3 y' - 3 \frac{y}{x} + \sqrt{1 - (\frac{y}{x})^2} = 0$. Let

Vedclass · Accepted Answer

$3 \cos^{-1}\left(\frac{y}{x}\right) = \ln |x|$

Vedclass · Answer

(A) Given differential equation: $3 x y' - 3 y + \sqrt{x^2 - y^2} = 0$. Dividing by $x$: $3 y' - 3 \frac{y}{x} + \sqrt{1 - (\frac{y}{x})^2} = 0$. Let $y = vx$, then $y' = v + x \frac{dv}{dx}$. Substituting: $3(v + x \frac{dv}{dx}) - 3v + \sqrt{1 - v^2} = 0$. $3v + 3x \frac{dv}{dx} - 3v + \sqrt{1 - v^2} = 0$. $3x \frac{dv}{dx} = -\sqrt{1 - v^2}$. Separating variables: $\frac{-dv}{\sqrt{1 - v^2}} = \frac{1}{3} \frac{dx}{x}$. Integrating both sides: $\int \frac{-dv}{\sqrt{1 - v^2}} = \int \frac{1}{3} \frac{dx}{x}$. $\cos^{-1}(v) = \frac{1}{3} \ln |x| + C$. Since $y(1) = 1$, $v = \frac{y}{x} = 1$ when $x = 1$. $\cos^{-1}(1) = \frac{1}{3} \ln(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$. Thus, $\cos^{-1}(\frac{y}{x}) = \frac{1}{3} \ln |x|$, which implies $3 \cos^{-1}(\frac{y}{x}) = \ln |x|$.

The solution of the differential equation $3 x y' - 3 y + (x^2 - y^2)^{1/2} = 0$, satisfying the condition $y(1) = 1$ is

Similar Questions

The general solution of $\frac{dy}{dx} = \frac{x+y}{x-y}$ is

The solution of $(x^2+y^2) dx = 2xy dy$ is:

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'

Identify the statement$(s)$ which is/are True.

Prove that $x^{2}-y^{2}=c(x^{2}+y^{2})^{2}$ is the general solution of the differential equation $(x^{3}-3xy^{2})dx=(y^{3}-3x^{2}y)dy,$ where $c$ is a parameter.

Vedclass Test Series

Exam Paper Generator

Online Exam Module