Let y^{prime}(x) + y(x) g^{prime}(x) = g(x) g | vedclass

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

Question

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = g^{\prime}(x)$ and $Q(x

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = g^{\prime}(x)$ and $Q(x) = g(x)g^{\prime}(x)$.The integrating factor ($I$.$F$.) is given by $e^{\int P(x) dx} = e^{\int g^{\prime}(x) dx} = e^{g(x)}$.The general solution is $y \cdot e^{g(x)} = \int Q(x) e^{g(x)} dx + C = \int g(x) g^{\prime}(x) e^{g(x)} dx + C$.Let $u = g(x)$,then $du = g^{\prime}(x) dx$. The integral becomes $\int u e^u du = u e^u - e^u = e^{g(x)}(g(x) - 1)$.Thus,$y e^{g(x)} = e^{g(x)}(g(x) - 1) + C$.Given $y(0) = 0$ and $g(0) = 0$,we substitute these values: $0 \cdot e^0 = e^0(0 - 1) + C \implies 0 = -1 + C \implies C = 1$.So,the solution is $y e^{g(x)} = e^{g(x)}(g(x) - 1) + 1$.To find $y(2)$,substitute $x = 2$ and $g(2) = 0$: $y(2) e^0 = e^0(0 - 1) + 1 \implies y(2) = -1 + 1 = 0$.

Let $y^{\prime}(x) + y(x) g^{\prime}(x) = g(x) g^{\prime}(x)$,$y(0) = 0$,$x \in \mathbb{R}$,where $f^{\prime}(x)$ denotes $\frac{d f(x)}{d x}$ and $g(x)$ is a given non-constant differentiable function on $\mathbb{R}$ with $g(0) = g(2) = 0$. Then the value of $y(2)$ is

Similar Questions

The solution of the differential equation $\frac{dy}{dx} - 2y \tan 2x = e^x \sec 2x$ is

The integrating factor of the differential equation $\frac{dy}{dx}(1+x) - xy = 1-x$ is . . . . . . .

Let $y=f(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^6+4x}{\sqrt{1-x^2}}$ for $-1 < x < 1$ such that $f(0)=0$. If $6 \int_{-1/2}^{1/2} f(x) dx = 2\pi - \alpha$,then $\alpha^2$ is equal to . . . . . .

Let $y=y(x)$ be the solution of the differential equation $(1-x^2) dy = [xy + (x^3+2) \sqrt{3(1-x^2)}] dx$ for $-1 < x < 1$,with $y(0)=0$. If $y(1/2) = m/n$,where $m$ and $n$ are coprime numbers,then $m+n$ is equal to . . . . . . . . . .

The integrating factor of the differential equation $(x^2 + 1)\frac{dy}{dx} + 2xy = x^2 - 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module