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logistic_guy

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\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \frac{dy}{dx} = \frac{xy^2 - \cos x \sin x}{y(1 - x^2)}, \ \ \ y(0) = 2\)
 
\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \frac{dy}{dx} = \frac{xy^2 - \cos x \sin x}{y(1 - x^2)}, \ \ \ y(0) = 2\)
Please show us what you have tried and exactly where you are stuck.

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\(\displaystyle \frac{dy}{dx} = \frac{xy^2 - \cos x \sin x}{y(1 - x^2)}\)


\(\displaystyle (xy^2 - \cos x \sin x) \ dx + y(x^2 - 1) \ dy = 0\)


\(\displaystyle M = xy^2 - \cos x \sin x\)


\(\displaystyle N = y(x^2 - 1)\)


\(\displaystyle \frac{\partial M}{\partial y} = 2xy = \frac{\partial N}{\partial x}\)


\(\displaystyle \frac{\partial f}{\partial x} = M = xy^2 - \cos x \sin x\)


\(\displaystyle \int \frac{\partial f}{\partial x} \partial x = \int (xy^2 - \cos x \sin x) \ dx\)


\(\displaystyle f(x,y) = \frac{x^2y^2}{2} + \frac{\cos^2 x}{2} + g(y)\)


\(\displaystyle \frac{\partial f}{\partial y} = N = y(x^2 - 1) = x^2y + g'(y)\)


\(\displaystyle g'(y) = y(x^2 - 1) - x^2y\)


\(\displaystyle \frac{dg}{dy} = y(x^2 - 1) - x^2y\)


\(\displaystyle dg = (y(x^2 - 1) - x^2y) \ dy\)


\(\displaystyle \int dg = \int (y(x^2 - 1) - x^2y) \ dy\)


\(\displaystyle g(y) = \frac{y^2(x^2 - 1)}{2} - \frac{x^2y^2}{2}\)


\(\displaystyle f(x,y) = \frac{x^2y^2}{2} + \frac{\cos^2 x}{2} + \frac{y^2(x^2 - 1)}{2} - \frac{x^2y^2}{2} = \frac{\cos^2 x}{2} + \frac{y^2(x^2 - 1)}{2}\)

Then the solution to the differential equation is:

\(\displaystyle \frac{\cos^2 x}{2} + \frac{y^2(x^2 - 1)}{2} = D\)

Or

\(\displaystyle \cos^2 x + y^2(x^2 - 1) = 2D\)

Or

\(\displaystyle \cos^2 x + y^2(x^2 - 1) = C\)
 
y(0) = 2[/tex]
Let us apply the condition \(\displaystyle y(0) =2\) to the solution.

\(\displaystyle 1 + 2^2(- 1) = C\)

This gives:

\(\displaystyle C = -3\)

Then, the general solution becomes

\(\displaystyle \cos^2 x + y^2(x^2 - 1) = -3\)

Or

\(\displaystyle y^2(1 - x^2) - \cos^2 x = 3\)
 
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