Let y=y(x) be a differentiable function in the interval (0, ) such that y(1)=2, and _t x((…

Question

Let $y=y(x)$ be a differentiable function in the interval $(0, \infty)$ such that $y(1)=2$, and $\lim\limits_{t \rightarrow x}\left(\frac{t^2 y(x)-x^2 y(t)}{x-t}\right)=3$ for each $x > 0$. Then $2 y(2)$ is equal to :

Accepted Answer

$\underset{t	o x}{lim}\frac{{t}^{2}y(x)-{x}^{2}y(t)}{x-t}=3$

Substituting

$t	o x+h\Rightarrow h	o 0$

$\underset{h	o 0}{lim}\frac{(x+h{)}^{2}y(x)-{x}^{2}y(x+h)}{-h}=3$

${f}^{'}(x)=\underset{h	o 0}{lim}\frac{f(x+h)-f(x)}{h}$

$\Rightarrow$ $\underset{h	o 0}{lim}\frac{{x}^{2}y(x)+{h}^{2}y(x)+2xhy(x)-{x}^{2}y(x+h)}{-h}=3$

$\Rightarrow$ $\underset{h	o 0}{lim}\frac{-{x}^{2}[y(x+h)-y(x)]+y(x)({h}^{2}+2xh)}{-h}=3$

$\Rightarrow$ $\underset{h	o 0}{lim}\frac{{x}^{2}[y(x+h)-y(x)]}{h}-\underset{h	o 0}{lim}\frac{y(x)(h+2x)h}{h}=3$

Use

${f}^{'}(x)=\underset{h	o 0}{lim}\frac{f(x+h)-f(x)}{h}$

${x}^{2}{y}^{'}(x)-2xy(x)=3$

Equation (1) is differential equation can be written by replacing

${y}^{'}(x)=\frac{dy}{dx}\, \, y(x)=y$

$\Rightarrow$ ${x}^{2}\frac{dy}{dx}-2xy=3$

$\Rightarrow$ $\frac{dy}{dx}-\frac{2}{x}y=\frac{3}{{x}^{2}}$ (Linear differential equation)

Integrating factor

$={e}^{\int -\frac{2}{x}dx}={e}^{-2\ln x}={e}^{\ln (\frac{1}{{x}^{2}})}=\frac{1}{{x}^{2}}$

$	herefore $ Solution of linear differential equation

$y\cdot ()=\int ()\cdot \frac{3}{{x}^{2}}dx$

$\Rightarrow$ $y\cdot \frac{1}{{x}^{2}}=\int \frac{1}{{x}^{2}}\cdot \frac{3}{{x}^{2}}dx=\int \frac{3}{{x}^{4}}dx$

$\Rightarrow$ $\frac{y}{{x}^{2}}=3	imes \frac{{x}^{-4+1}}{-4+1}=-\frac{1}{{x}^{3}}+C$

$\Rightarrow$ $y=-\frac{1}{x}+C{x}^{2}$ (General solution)

Apply initial condition

$y(1)=2$

$\Rightarrow$$2=-\frac{1}{1}+C(1)\Rightarrow C=3$

So particular solution is : $y(x)=-\frac{1}{x}+3{x}^{2}$

$	herefore $ $y(2)=-\frac{1}{2}+3(2{)}^{2}=12-\frac{1}{2}=\frac{23}{2}$

$\Rightarrow$ $2y(2)=23$

Solutions · 1