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Degger
4 days ago
12

1. X^4(dy/dx) +x^3y =- sec (xy)Integral by separation of variables? ​

Mathematics
1 answer:
PIT_PIT [3.9K]4 days ago
8 0

Answer:

Step-by-step explanation:

Considering the differential equation x^4(dy/dx) + x^3y = -sec(xy). We will solve it employing the method of separation of variables;

x^{4} \frac{dy}{dx} +x^{3}y = -sec(xy)\\x^{3}(x\frac{dy}{dx} + y) = -sec(xy)\\let \ v=xy\\\frac{dv}{dx} = x\frac{dy}{dx} + y(implicit \ in\ nature)\\

By substituting v and dv/dx into the previous equation, we acquire;

x^{3}\frac{dv}{dx} = -secv

We then separate the variables:

-\frac{dv}{secv} = \frac{dx}{x^{3} }

-cosvdv = x^{-3}dx\\ integrating\ both\ sides\\-\int\limits {cosv} \, dv = \int\limits {x^{-3} } \, dx\\-sinv = \frac{x^{-2} }{-2} + C\\since\ v = xy\\-sinxy = \frac{x^{-2} }{-2} + C\\2sin(xy) = x^{-2} -2C\\2 sin(xy) = \frac{1}{x^{2} } -K (where\ K = 2C)\\

The end expression provides the solution to the differential equation.

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