The general solution of the differential equa | vedclass

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

Question

(B) Given differential equation: $(y^2 - x^3) dx - xy dy = 0$Divide by $dx$ (assuming $dx 
eq 0$):$y^2 - x^3 - xy \frac{dy}{dx} = 0$$xy \frac{dy}{dx}

Vedclass · Accepted Answer

$y^2 - 2x^3 + cx^2 = 0$

Vedclass · Answer

(B) Given differential equation: $(y^2 - x^3) dx - xy dy = 0$Divide by $dx$ (assuming $dx 
eq 0$):$y^2 - x^3 - xy \frac{dy}{dx} = 0$$xy \frac{dy}{dx} - y^2 = -x^3$Divide by $x$:$y \frac{dy}{dx} - \frac{1}{x} y^2 = -x^2$  ......$(i)$Let $y^2 = v$,then $2y \frac{dy}{dx} = \frac{dv}{dx}$,so $y \frac{dy}{dx} = \frac{1}{2} \frac{dv}{dx}$.Substituting into $(i)$:$\frac{1}{2} \frac{dv}{dx} - \frac{1}{x} v = -x^2$$\frac{dv}{dx} - \frac{2}{x} v = -2x^2$  ......$(ii)$This is a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = -2x^2$.Integrating factor $I.F. = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln |x|} = x^{-2} = \frac{1}{x^2}$.The solution is $v \cdot I.F. = \int Q(x) \cdot I.F. dx + c$.$v \cdot \frac{1}{x^2} = \int (-2x^2) \cdot \frac{1}{x^2} dx + c$$\frac{v}{x^2} = \int -2 dx + c$$\frac{v}{x^2} = -2x + c$Since $v = y^2$,we have $\frac{y^2}{x^2} = -2x + c$.$y^2 = -2x^3 + cx^2$,which implies $y^2 - 2x^3 + cx^2 = 0$.Thus,the correct option is $B$.

The general solution of the differential equation $(y^2 - x^3) dx - xy dy = 0, (x \neq 0)$ is (where $c$ is a constant of integration)

Similar Questions

The solution of the differential equation,$x^2 \frac{dy}{dx} \cos \frac{1}{x} - y \sin \frac{1}{x} = -1,$ where $y \rightarrow -1$ as $x \rightarrow \infty$ is

Let $y=y(x)$ be the solution of the differential equation $(x-x^{3}) dy=(y+yx^{2}-3x^{4}) dx, x>2$. If $y(3)=3$,then $y(4)$ is equal to :

If a curve $y = f(x)$ passes through the point $(1, 2)$ and satisfies $x \frac{dy}{dx} + y = bx^4$,then for what value of $b$ is $\int_{1}^{2} f(x) dx = \frac{62}{5}$?

Let the solution curve $y=f(x)$ of the differential equation $\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{4}+2x}{\sqrt{1-x^{2}}}, x \in(-1,1)$ pass through the origin. Then $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$ is equal to

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=2(y+2 \sin x-5)x-2 \cos x$ such that $y(0)=7$. Then $y(\pi)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module