1.Find value of 𝛉 in Mean Value theorem f(x+h)= f(x)+hf^( ′) (x+θh), if f(x)=(1/x) 2.If (√(1−x^2 )) +(√(1−y^2 )) = k(x−y) prove that (dy/dx) = ((√(1−y^2 ))/(√(1−x^2 ))). 3. Solve the differential equation: x^2 (d^2 y/dx^2 )−3x(dy/dx)+3y = x^2 (2x−1).


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Question Number 94418 by niroj last updated on 18/May/20

$1 . F ind value of θ in Mean Value theorem$ $f (x + h) = f (x) + {hf}^{^{'}} (x + θ h), if f (x) = \frac{1}{x}$ $2 . If \sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = k (x - y) prove that$ $\frac{dy}{dx} = \frac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}} .$ $3 . Solve the differential equation :$ $x^{2} \frac{d^{2} y}{{dx}^{2}} - 3 x \frac{dy}{dx} + 3 y = x^{2} (2 x - 1) .$
	Answered by som(math1967) last updated on 18/May/20

	$1 . f (x) = \frac{1}{x} ∴ f^{^{'}} (x) = - \frac{1}{x^{2}}$ $now f (x + h) = f (x) + {hf}^{^{'}} (x + θ h)$ $\frac{1}{(x + h)} = \frac{1}{x} - \frac{h}{{(x + θ h)}^{2}}$ $\frac{- h}{x (x + h)} = \frac{- h}{{(x + θ h)}^{2}}$ ${(x + θ h)}^{2} = x^{2} + hx$ $(x + θ h) = \pm \sqrt{x^{2} + hx}$ $θ = \frac{\pm \sqrt{x^{2} + hx} - x}{h} ans$
	Commented by niroj last updated on 18/May/20
	nice ��thank you.
	Answered by som(math1967) last updated on 18/May/20

	$2 . let x = \sin α y = \sin β$ $∴ \cos α + \cos β = k (\sin α - \sin β)$ $2 \cos \frac{α + β}{2} \cos \frac{α - β}{2} = 2 ksin \frac{α - β}{2} \cos \frac{α + β}{2}$ $\frac{\cos \frac{α - β}{2}}{\sin \frac{α - β}{2}} = k$ $\frac{α - β}{2} = \cot^{- 1} k$ $α - β = {2 \cot}^{- 1} k$ $\sin^{- 1} x - \sin^{- 1} y = {2 \cot}^{- 1} k$ $\frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - y^{2}}} . \frac{dy}{dx} = 0$ $∴ \frac{dy}{dx} = \frac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}}$
	Commented by niroj last updated on 18/May/20
	superb�� thank dear for your effort.
	Answered by Mr.D.N. last updated on 18/May/20

	$3 . x^{2} \frac{d^{2} y}{{dx}^{2}} - 3 x \frac{dy}{dx} + 3 y = x^{2} (2 x - 1)$ $Second order linear differential {eq}^{n}$ $with variable coefficient$ $\frac{d^{2} y}{{dx}^{2}} - \frac{3}{x} \frac{dy}{dx} + \frac{3}{x^{2}} y = 2 x - 1 . . . . . (i)$ $P = - \frac{3}{x}, Q = \frac{3}{x^{2}} and R = 2 x - 1$ $If P + Qx = 0 then Part of CF u = x$ $- \frac{3}{x} + \frac{3}{x^{2}} x = 0$ $for complete {sol}^{n} :$ $y = uv$ $y = vx . . . . . . . (ii)$ $We know the form :$ $\frac{d^{2} v}{{dx}^{2}} + (P + \frac{2}{u} . \frac{du}{dx}) \frac{dv}{dx} = \frac{R}{u}$ $\frac{d^{2} v}{{dx}^{2}} + (- \frac{3}{x} + \frac{2}{x} . \frac{dx}{dx}) \frac{dv}{dx} = \frac{2 x - 1}{x}$ $\frac{d^{2} v}{{dx}^{2}} - \frac{1}{x} . \frac{dv}{dx} = \frac{2 x - 1}{x}$ $Change to linear form,$ $Put, \frac{dv}{dx} = t \Rightarrow \frac{d^{2} v}{{dx}^{2}} = \frac{dt}{dx}$ $\frac{dt}{dx} - \frac{1}{x} t = \frac{2 x - 1}{x}$ $Now, IF = e^{\int Pdx} = e^{- \int \frac{1}{x} dx} = e^{- \log x}$ $IF = e^{\log x^{- 1}} = \frac{1}{x}$ $t . \frac{1}{x} = \int \frac{1}{x} . \frac{2 x - 1}{x} dx + C_{1}$ $\frac{t}{x} = \int (\frac{2}{x} - x^{- 2}) dx + C_{1}$ $t . \frac{1}{x} = 2 \log x - \frac{x^{- 1}}{(- 1)} + C_{1}$ $t . \frac{1}{x} = 2 \log x + \frac{1}{x} + C_{1}$ $t = 2 xlog x + 1 + C_{1} x$ $\frac{dv}{dx} = 2 x \log x + 1 + C_{1} x$ $\int dv = 2 \int x \log x dx + \int dx + C_{1} \int xdx + C_{2}$ $v = 2 [\log x . \frac{x^{2}}{2} - \int \frac{1}{x} . \frac{x^{2}}{2} dx] + x + \frac{x^{2}}{2} C_{1} + C_{2}$ $v = 2 . \frac{x^{2}}{2} \log x - 2 . \frac{1}{2} \int xdx + x + \frac{x^{2}}{2} C_{1} + C_{2}$ $v = x^{2} \log x - \frac{x^{2}}{2} + x + \frac{x^{2}}{2} C_{1} + C_{2}$ $Now again Putting the value of v in {equ}^{n} (ii)$ $y = vx$ $y = x^{3} \log x - \frac{x^{3}}{2} + x^{2} + \frac{x^{3}}{2} C_{1} + C_{2} x .$
	Commented by niroj last updated on 18/May/20
	it's outstanding��
	Commented by peter frank last updated on 24/May/20

	$good$
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