Question

1. X^4(dy/dx) +x^3y =- sec (xy)

Integral by separation of variables?
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Answer 1

Answer:

Step-by-step explanation:

given the differential equation x^4(dy/dx) +x^3y =- sec (xy). Solving using the variable separable method;

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

Substituting v and dv/dx into the equation above we have;

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

Separating 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 final expression gives the solution to the differential equation.