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Question: The solution of \(\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^{2}}(\log z)^{2}\) is...

The solution of dzdx+zxlog⁡z=zx2(log⁡z)2\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^{2}}(\log z)^{2} is

A

(1log⁡z)x=2−x2c\left( \frac{1}{\log z} \right)x = 2 - x^{2}c

B

(1log⁡z)x=2+x2c\left( \frac{1}{\log z} \right)x = 2 + x^{2}c

C

(1log⁡z)x=x2c\left( \frac{1}{\log z} \right)x = x^{2}c

D

(1log⁡z)x=12+cx2\left( \frac{1}{\log z} \right)x = \frac{1}{2} + cx^{2}`

Answer

(1log⁡z)x=12+cx2\left( \frac{1}{\log z} \right)x = \frac{1}{2} + cx^{2}`

Explanation

Solution

Dividing the given equation by z(log⁡z)2z(\log z)^{2},

1z(log⁡z)2dzdx+1x1log⁡z=1x2\frac{1}{z(\log z)^{2}}\frac{dz}{dx} + \frac{1}{x}\frac{1}{\log z} = \frac{1}{x^{2}}

Let 1log⁡z=t\frac{1}{\log z} = t ⇒ −1(log⁡z)2⋅12dzdx=dtdx- \frac{1}{(\log z)^{2}} \cdot \frac{1}{2}\frac{dz}{dx} = \frac{dt}{dx}

∴ −dtdx+tx=1x2- \frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^{2}}

⇒ dtdx−tx=−1x2\frac{dt}{dx} - \frac{t}{x} = - \frac{1}{x^{2}} ……..(i)

I.F.=e∫−dxx=e−ln⁡x=eln⁡1x=1x= e^{\int_{}^{}{- \frac{dx}{x}}} = e^{- \ln x} = e^{\ln\frac{1}{x}} = \frac{1}{x}

Multiplying (i) by 1x\frac{1}{x} and integrating,

tx=∫−1x3dx=12x2+c\frac{t}{x} = \int_{}^{}{- \frac{1}{x^{3}}dx = \frac{1}{2x^{2}} + c} ⇒ 1xlog⁡z=12x2+c\frac{1}{x\log z} = \frac{1}{2x^{2}} + c

∴ (1ln⁡z)x=(12)+cx2\left( \frac{1}{\ln z} \right)x = \left( \frac{1}{2} \right) + cx^{2}