Please help me solve this Differential Equation
2ydx+x(x2lny -1)dy=0
Differential Equation $2y \, dx+x(x^2 \ln y -1) \, dy = 0$
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2ydx+x (x²lny-1)dy=0
$2y \, dx+x(x^2 \ln y -1) \, dy = 0$
$dx + x^3 \left( \dfrac{\ln y}{y} \right) \, dy - \dfrac{x}{y} \, dy = 0$
$dx - \dfrac{1}{y} \, x \, dy = -\dfrac{\ln y}{y} \, x^3 \, dy$ ← Bernoulli's Equation
$dx + P(y) \, x \, dy = Q(y) \, x^n \, dy$.
Where:
$P(y) = -\dfrac{1}{y}$
$Q(y) = -\dfrac{\ln y}{y}$
$n = 3$
Integrating factor
$\displaystyle u = e^{(1 - n) \int P(y) \, dy} = e^{-2\int (-1/y) \, dy} = e^{2\int (dy/y)}$
$u = e^{2\ln y} = e^{\ln y^2} = y^2$
The solution to Bernoulli's Equation is:
$\displaystyle x^{1 - n} u = (1 - n) \int Q(y) \, u \, dy + C$
Hence,
$\displaystyle x^{-2} y^2 = -2 \int \left( -\dfrac{\ln y}{y} \right)(y^2) \, dy + C$
$\displaystyle x^{-2} y^2 = 2 \int y \ln y \, dy + C$
$du = \dfrac{dy}{y}$
$v = \frac{1}{2} y^2$
Therefore,
${\displaystyle \int} y \ln y \, dy = \frac{1}{2}y^2 \ln y - {\displaystyle \int} \frac{1}{2} y^2 \left( \dfrac{dy}{y} \right)$
${\displaystyle \int} y \ln y \, dy = \frac{1}{2}y^2 \ln y - \frac{1}{2} {\displaystyle \int} y \, dy$
${\displaystyle \int} y \ln y \, dy = \frac{1}{2}y^2 \ln y - \frac{1}{4}y^2$
Thus,
$x^{-2} y^2 = 2 \left( \frac{1}{2}y^2 \ln y - \frac{1}{4}y^2 \right) + C$
$x^{-2} y^2 = y^2 \ln y - \frac{1}{2}y^2 + C$
$y^2 \left( \dfrac{1}{x^2} - \ln y + \dfrac{1}{2} \right) = C$ ← answer
Kindly double-check my solution.
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