The solution of the differential equation x , | vedclass

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

Question

(B) Given differential equation: $x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx$Divide by $dx$: $x \frac{dy}{dx} - y = \sqrt{x^2 + y^2}$$x \frac{dy}{dx}

Vedclass · Accepted Answer

$y + \sqrt{x^2 + y^2} = cx^2$

Vedclass · Answer

(B) Given differential equation: $x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx$Divide by $dx$: $x \frac{dy}{dx} - y = \sqrt{x^2 + y^2}$$x \frac{dy}{dx} = y + \sqrt{x^2 + y^2}$$\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation:$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2x^2}}{x} = v + \sqrt{1 + v^2}$$x \frac{dv}{dx} = \sqrt{1 + v^2}$Separating variables: $\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$Integrating both sides: $\ln|v + \sqrt{1 + v^2}| = \ln|x| + \ln|c|$$v + \sqrt{1 + v^2} = cx$Substitute $v = \frac{y}{x}$:$\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = cx$$\frac{y + \sqrt{x^2 + y^2}}{x} = cx$$y + \sqrt{x^2 + y^2} = cx^2$

The solution of the differential equation $x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx$ is

Similar Questions

If $\frac{dy}{dx} = \frac{y + x \tan(\frac{y}{x})}{x}$, then $\sin(\frac{y}{x})$ is equal to

Solve the differential equation $y e^{\frac{x}{y}} dx = \left( x e^{\frac{x}{y}} + y^2 \right) dy$ where $y \neq 0$.

The solution curve of the differential equation $y \frac{dx}{dy} = x(\log_e x - \log_e y + 1)$,$x > 0, y > 0$ passing through the point $(e, 1)$ is

$A$ bag contains $5$ red balls and $3$ green balls. $A$ ball is selected at random and not replaced. $A$ second ball is then selected. The probability of selecting one red ball and one green ball is

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