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Math Help - Linear and separable equations

  1. #1
    Junior Member
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    Linear and separable equations

    Can someone help me with these, i've been looking through my University maths book but couldn't figure out how to do any of these. Any help would be great:

    First one

    Solve:

    Dy/Dx= e^y .Sin^2 X/SecX

    X^2 Dy/Dx +2xy=Cos^2X

    2 D^2y/Dx^2 -5 dy/dx +2y=0
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  2. #2
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    Hello, Mathsnewbie!

    You should have been taught to recognize and solve different forms.


    \frac{dy}{dx}\:=\:\frac{e^y\sin^2\!x}{\sec x}
    Separate variables: . e^{-y}dy \:=\:\sin^2\!x\cos x\,dx

    Integrate: . -e^{-y} \:=\:\tfrac{1}{3}\sin^3\!x + C . . . etc.



    x^2\,\frac{dy}{dx} +2xy\:=\:\cos^2\!x
    This requires an Integrating Factor . . .

    Divide by x^2\!:\;\;\frac{dy}{dx} + \frac{2}{x}\,y \:=\:\frac{\cos^2\!x}{x^2}

    . . Then: . I \;=\;e^{\int\frac{2}{x}dx} \;=\;e^{2\ln x} \;=\;e^{\ln(x^2)} \;=\;x^2

    Multiply by I\!:\;\;x^2\,\frac{dy}{dx} + 2xy \:=\:\cos^2\!x

    And we have: . \frac{d}{dx}\left(x^2y\right) \:=\:\cos^2\!x

    Integrate: . x^2y \:=\:\int\cos^2\!x\,dx . . . etc.



    2\,\frac{d^2y}{dx^2} -5\,\frac{dy}{dx} +2y\:=\:0

    The characteristic equation is: . 2m^2 - 5m + 2 \:=\:0

    . . which factors: . (m-2)(2m-1) \:=\:0

    . . and has roots: . m \:=\:2,\tfrac{1}{2}

    Therefore: . y \:=\:C_1e^{2x} + C_2e^{\frac{1}{2}x}

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