The general solution of the differential equa | vedclass

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

Question

(A) Given the differential equation: $\frac{dy}{dx} + xy = 4x - 2y + 8$. Rearranging the terms to standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:

Vedclass · Accepted Answer

$y = 4 + ce^{-\frac{x^2}{2} - 2x}$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} + xy = 4x - 2y + 8$. Rearranging the terms to standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$: $\frac{dy}{dx} + (x + 2)y = 4x + 8$. $\frac{dy}{dx} + (x + 2)y = 4(x + 2)$. Here,$P(x) = x + 2$ and $Q(x) = 4(x + 2)$. The integrating factor $IF = e^{\int P(x) dx} = e^{\int (x + 2) dx} = e^{\frac{x^2}{2} + 2x}$. The general solution is given by $y \cdot IF = \int Q(x) \cdot IF dx + c$. $y \cdot e^{\frac{x^2}{2} + 2x} = \int 4(x + 2) e^{\frac{x^2}{2} + 2x} dx + c$. Let $u = \frac{x^2}{2} + 2x$,then $du = (x + 2) dx$. $y \cdot e^{\frac{x^2}{2} + 2x} = 4 \int e^u du + c = 4e^u + c = 4e^{\frac{x^2}{2} + 2x} + c$. Dividing both sides by $e^{\frac{x^2}{2} + 2x}$: $y = 4 + ce^{-(\frac{x^2}{2} + 2x)}$. Thus,the correct option is $A$.

The general solution of the differential equation $\frac{dy}{dx} + xy = 4x - 2y + 8$ is

Similar Questions

Let $f: [1, \infty) \to \mathbb{R}$ be a differentiable function defined as $f(x) = \int_1^x f(t) \, dt + (1 - x)(\log_e x - 1) + e$. Then the value of $f(f(1))$ is:

The general solution of the differential equation $(\sin y \cos^2 y - x \sec^2 y) dy = (\tan y) dx$ is

Solve the differential equation $\left(\tan ^{-1} y-x\right) d y=\left(1+y^{2}\right) d x$.

Let $y(x)$ be a solution of the differential equation $(1+e^x) y^{\prime}+y e^x=1$. If $y(0)=2$,then which of the following statements is (are) true?
$(A)$ $y(-4)=0$
$(B)$ $y(-2)=0$
$(C)$ $y(x)$ has a critical point in the interval $(-1,0)$
$(D)$ $y(x)$ has no critical point in the interval $(-1,0)$

If $y=f(x)$ is the solution of the differential equation $(1+\cos^2 x) f'(x) - f(x) \sin 2x = 4 \sin 2x$ with $f(0)=0$,then $f(\frac{\pi}{3})=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module