Let f :(0, infty) rightarrow R be a function | vedclass

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

Question

(C) Given the differential equation: $x^2 f^{\prime}(x) - 2 x f(x) = 3$.Divide both sides by $x^4$ (where $x 
eq 0$):$\frac{x^2 f^{\prime}(x) - 2 x f

Vedclass · Accepted Answer

$39$

Vedclass · Answer

(C) Given the differential equation: $x^2 f^{\prime}(x) - 2 x f(x) = 3$.Divide both sides by $x^4$ (where $x 
eq 0$):$\frac{x^2 f^{\prime}(x) - 2 x f(x)}{x^4} = \frac{3}{x^4}$This simplifies to the derivative of a quotient:$\frac{d}{dx} \left( \frac{f(x)}{x^2} ight) = 3 x^{-4}$Integrating both sides with respect to $x$:$\frac{f(x)}{x^2} = \int 3 x^{-4} dx = 3 \left( \frac{x^{-3}}{-3} ight) + C = -\frac{1}{x^3} + C$Multiplying by $x^2$:$f(x) = -\frac{1}{x} + C x^2$Given $f(1) = 4$,substitute $x=1$:$4 = -\frac{1}{1} + C(1)^2 \Rightarrow 4 = -1 + C \Rightarrow C = 5$.Thus,$f(x) = 5x^2 - \frac{1}{x}$.Now,calculate $2f(2)$:$f(2) = 5(2)^2 - \frac{1}{2} = 20 - 0.5 = 19.5$.$2f(2) = 2 	imes 19.5 = 39$.

Let $f :(0, \infty) \rightarrow R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2 f^{\prime}(x)=2 x f(x)+3$,with $f(1)=4$. Then $2 f(2)$ is equal to:

Similar Questions

The general solution of the differential equation $(x+2y^3) \frac{dy}{dx} - y = 0, y > 0$ is

If $x \log x \frac{dy}{dx} + y = \log x^2$ and $y(e) = 0$,then $y(e^2) = $

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 :

The solution of the differential equation $\frac{dy}{dx} - 2\frac{y}{x} = x^3$ is:

If $\frac{dx}{dy} = \frac{1+x-y^2}{y}$ and $x(1) = 1$,then $5x(2)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module