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Thread: Differential equations substitution

  1. #1
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    Differential equations substitution

    Solve:

    $\displaystyle xyy'+y^2=sinx$ by letting $\displaystyle y^2=v$

    Would you use this and then substitute it into the differential equation?

    $\displaystyle y'=\dfrac{1}{2\sqrt{v}}\dfrac{dv}{dx}$
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  2. #2
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    $\displaystyle y^2 = v$

    $\displaystyle y = \sqrt{v}$

    $\displaystyle \frac{d}{dx}(y) = \frac{d}{dx}(\sqrt{v})$

    $\displaystyle \frac{dy}{dx} = \frac{d}{dv}(\sqrt{v})\,\frac{dv}{dx}$

    $\displaystyle \frac{dy}{dx} = \frac{1}{2\sqrt{v}}\,\frac{dv}{dx}$.


    So yes, I would substitute this into the DE...

    $\displaystyle x\sqrt{v}\left(\frac{1}{2\sqrt{v}}\,\frac{dv}{dx}\ right) + v = \sin{x}$

    $\displaystyle \frac{1}{2}x\,\frac{dv}{dx} + v = \sin{x}$

    $\displaystyle \frac{dv}{dx} + \frac{2}{x}v = \frac{2}{x}\sin{x}$


    Now using the integrating factor

    $\displaystyle e^{\int{\frac{2}{x}\,dv}} = e^{2\ln{x}} = e^{\ln{(x^2)}} = x^2$ and multiplying through gives

    $\displaystyle x^2\,\frac{dv}{dx} + 2xv = 2x\sin{x}$

    $\displaystyle \frac{d}{dx}(x^2v) = 2x\sin{x}$

    $\displaystyle x^2v = 2\int{x\sin{x}\,dx}$

    $\displaystyle x^2v = 2\left[-x\cos{x} - \int{\cos{x}\,dx}\right]$

    $\displaystyle x^2v = 2\left[-x\cos{x} - \sin{x}\right] + C$

    $\displaystyle x^2v = -2x\cos{x} - 2\sin{x} + C$

    $\displaystyle v = -\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}$

    $\displaystyle y^2 = -\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}$

    $\displaystyle y = \pm\sqrt{-\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}}$.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    $\displaystyle y^2 = v$

    $\displaystyle y = \sqrt{v}$

    $\displaystyle \frac{d}{dx}(y) = \frac{d}{dx}(\sqrt{v})$

    $\displaystyle \frac{dy}{dx} = \frac{d}{dv}(\sqrt{v})\,\frac{dv}{dx}$

    $\displaystyle \frac{dy}{dx} = \frac{1}{2\sqrt{v}}\,\frac{dv}{dx}$.


    So yes, I would substitute this into the DE...

    $\displaystyle x\sqrt{v}\left(\frac{1}{2\sqrt{v}}\,\frac{dv}{dx}\ right) + v = \sin{x}$

    $\displaystyle \frac{1}{2}x\,\frac{dv}{dx} + v = \sin{x}$

    $\displaystyle \frac{dv}{dx} + \frac{2}{x}v = \frac{2}{x}\sin{x}$


    Now using the integrating factor

    $\displaystyle e^{\int{\frac{2}{x}\,dv}} = e^{2\ln{x}} = e^{\ln{(x^2)}} = x^2$ and multiplying through gives

    $\displaystyle x^2\,\frac{dv}{dx} + 2xv = 2x\sin{x}$

    $\displaystyle \frac{d}{dx}(x^2v) = 2x\sin{x}$

    $\displaystyle x^2v = 2\int{x\sin{x}\,dx}$

    $\displaystyle x^2v = 2\left[-x\cos{x} - \int{\cos{x}\,dx}\right]$

    $\displaystyle x^2v = 2\left[-x\cos{x} - \sin{x}\right] + C$

    $\displaystyle x^2v = -2x\cos{x} - 2\sin{x} + C$

    $\displaystyle v = -\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}$

    $\displaystyle y^2 = -\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}$

    $\displaystyle y = \pm\sqrt{-\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}}$.
    Thanks, that makes a lot of sense now
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