The solution of the equation xfrac{dy}{dx} + | vedclass

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

Question

(B) Given the linear differential equation: $x\frac{dy}{dx} + 3y = x$. Divide the entire equation by $x$ to get it into the standard form $\frac{dy}{d

Vedclass · Accepted Answer

$x^3y = \frac{x^4}{4} + c$

Vedclass · Answer

(B) Given the linear differential equation: $x\frac{dy}{dx} + 3y = x$. Divide the entire equation by $x$ to get it into the standard form $\frac{dy}{dx} + Py = Q$: $\frac{dy}{dx} + \frac{3}{x}y = 1$. Here,$P = \frac{3}{x}$ and $Q = 1$. Calculate the Integrating Factor $(I.F.)$: $I.F. = e^{\int P dx} = e^{\int \frac{3}{x} dx} = e^{3 \ln|x|} = e^{\ln|x^3|} = x^3$. The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$. Substituting the values: $y \cdot x^3 = \int 1 \cdot x^3 dx + c$. $y x^3 = \frac{x^4}{4} + c$.

The solution of the equation $x\frac{dy}{dx} + 3y = x$ is

Similar Questions

Let $f:[1, \infty) \rightarrow \mathbb{R}$ be a differentiable function. If $6 \int_{1}^{x} f(t) dt = 3xf(x) + x^{3} - 4$ for all $x \ge 1$, then the value of $f(2) - f(3)$ is

The solution of the differential equation $y^{\prime} = \frac{1}{e^y - x}$ is

$A$ family of curves has the differential equation $x y \frac{d y}{d x}=2 y^2-x^2$. Then, the family of curves is

Which of the following equations is a linear differential equation?

The solution of the differential equation $\frac{dy}{dx} + y \cot x = 2 \cos x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module