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Math Help - Diff equation please

  1. #1
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    Diff equation please

    (a) 1/x * dy/dx + 2/(1+x^2)*y=3 y(1)=4


    (b) 2*x * dy/dx = 2*y + (x^2+y^2)^1/2

    Thank you very much
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  2. #2
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    I'll give you the solution to b, you try to work to that end. Okey-doke?.

    b): \frac{5}{x^{2}+1}+\frac{3(x^{2}+1)}{4}
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  3. #3
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    Quote Originally Posted by simfonija
    (a) 1/x * dy/dx + 2/(1+x^2)*y=3 y(1)=4
    \frac{1}{x}\frac{dy}{dx}+\frac{2}{1+x^2}y=3
    Multiply by "x",
    \frac{dy}{dx}+\frac{2x}{1+x^2}y=3x
    This is a Linear-First Order Differential Equation.
    Here,
    P(x)=\frac{2x}{1+x^2}
    Q(x)=3x
    Therefore, u=1+x^2
    I(x)=\exp \left( \int \frac{2x}{1+x^2} dx\right) =e^{\ln (1+x^2)}=1+x^2
    Thus, the solutions are,
    \frac{1}{I(x)}\cdot \int I(x)Q(x) dx
    Thus,
    (1+x^2)^{-1}\int \frac{3x}{1+x^2} dx
    Use, u=1+x^2,
    (1+x^2)^{-1}(3/2\ln (1+x^2)+C)
    Now you probably mean,
    y(0)=4 because these are usually how initial conditions are given.
    Thus,
    C=4
    From here the unique solution is,
    \frac{3\ln (1+x^2)}{2  (1+x^2)}+\frac{4}{1+x^2}
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