1-Find-value-of-in-Mean-Value-theorem-f-x-h-f-x-hf-x-h-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-d

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 \dots . . (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 \dots \dots . (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

good