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If xdy /dx= y (log y - log x + 1), then the solution of the equation is ?

sudhanshu , 11 Years ago
Grade 12
anser 2 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
x.\frac{dy}{dx} = y(logy-logx+1)
\frac{dy}{dx} = \frac{y}{x}(log\frac{y}{x}+1)
y = vx
\frac{dy}{dx} = v + x.\frac{dv}{dx}
v + x.\frac{dv}{dx} = v (logv + 1))
x.\frac{dv}{dx} = v logv
\frac{dv}{v.logv} = \frac{dx}{x}
\int \frac{dv}{v.logv} = \int \frac{dx}{x}
log(logv) = logx + constant
log(log\frac{y}{x}) = logx + constant
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the solution to your question.
 
Given, xdy /dx= y (log y - log x + 1)
→ dy/dx = y/x (log(y/x) + 1)
Let, y = vx
Hence, dy/dx = v + x.dv/dx
→ v + x.dv/dx = v (logv + 1)
→ x.dv/dx = v.logv
→ dv/v.logv = dx/x
Let, logv = t
Hence, dv/v = dt
→ dt/t = dx/x
Integrating both sides
→ ln(t) = ln(x) + c
or, t = C.x
→ log(y/x) = C.x
or, y/x = kx
or, y = x.kx
 
Hope it helps.
Thanks and regards,
Kushagra

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