Question: Solve $(1 + log xy) dx + (1 + \frac{x}{y}) dy = 0$
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Subject : Applied Mathematics 2

Topic : Linear Differential Equations With Constant Coefficients and Variable Coefficients Of Higher Order

Difficulty: High

mumbai university • 61 views
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modified 9 weeks ago  • written 9 weeks ago by gravatar for Ankit Pandey Ankit Pandey60
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Solution:

$\begin{array}{ll}{\text { Compare given eqn with }} & {\mathrm{Mdx}+\mathrm{Ndy}=0} \\ {\therefore \mathrm{M}=(1+\log \mathrm{x}, \mathrm{y})} & {\therefore \mathrm{N}=1+\frac{x}{y}} \\ {\frac{\partial M}{\partial y}=\frac{1}{x y} x=\frac{1}{y}} & {\frac{\partial N}{\partial x}=\frac{1}{y}}\end{array}$

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Hence the given differential eqn is exact.

The solution of exact differential eqn is given by,

$\int M d x+\int\left(N-\frac{\partial}{\partial y} \int M d x\right) d y=c$ -------------(1)

$\int M d x=\int(1+\log x y) d x=x+\log x y . x-x=x . \log x y$

$\frac{\partial}{\partial y} \int M d x=\frac{x}{y}$

$\int\left(N-\frac{\partial}{\partial y} \int M d x\right) \mathrm{dy}=\int\left(1+\frac{x}{y}-\frac{x} {y}\right) d y=y$

From eqn (1), the solution of given differential Equation is,

$x . \log x y+y=c$

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written 9 weeks ago by gravatar for Ankit Pandey Ankit Pandey60
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