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Math Help - Trig Substitution

  1. #1
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    Trig Substitution

    I don't know how to do this problem, so if someone could please help me, I'll be grateful:

    Solve the initial value problem

    (x2 + 1) squared dy/dx = the square root of(x2 + 1) , y(0) = 1

    Thanks
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  2. #2
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    Quote Originally Posted by turtle View Post
    I don't know how to do this problem, so if someone could please help me, I'll be grateful:

    Solve the initial value problem

    (x2 + 1) squared dy/dx = the square root of(x2 + 1) , y(0) = 1

    Thanks
    (x^2+1)y'=\sqrt{x^2+1}

    y'=\frac{\sqrt{x^2+1}}{x^2+1}=\frac{1}{\sqrt{1+x^2  }}

    y=\int \frac{1}{\sqrt{1+x^2}}dx=\sinh^{-1}(x)+C
    y(0)=1
    1=\sinh^{-1}(0)+C
    1=0+C
    C=1

    y=\sinh^{-1}(x)+1
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  3. #3
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    Hello, turtle!

    Your title was "Trig Substitution" . . .


    Solve the initial value problem:

    (x^2 + 1)^2\,\frac{dy}{dx} \:= \:\sqrt{x^2 + 1}\qquad y(0) = 1

    We have: . \frac{dy}{dx} \:=\:\frac{\sqrt{x^2+1}}{(x^2+1)^2}\quad\Rightarro  w\quad dy \:=\:\frac{dx}{(x^2+1)^{\frac{3}{2}}}

    Integrate: . \int dy \;=\;\int\frac{dx}{(x^2+1)^{\frac{3}{2}}}

    Let x = \tan\theta\quad\Rightarrow\quad dx = \sec^2\theta\,d\theta
    . . (x^2 + 1)^{\frac{3}{2}} \:=\:\left(\tan^2\theta + 1\right)^{\frac{3}{2}} \:=\:\left(\sec^2\theta\right)^{\frac{3}{2}} \:=\:\sec^3\theta

    Substitute: . y(x) \;=\;\int\frac{\sec^2\theta\,d\theta}{\sec^3\theta  } \;= \;\int\frac{d\theta}{\sec\theta} \;=\;\int\cos\theta\,d\theta \;=\;\sin\theta + C

    Back-substitute: . \tan\theta = x\quad\Rightarrow\quad\sin\theta = \frac{x}{\sqrt{x^2+1}}

    . . y(x) \;= \;\frac{x}{\sqrt{x^2+1}} + C

    Since y(0) = 1\!:\;\;\frac{0}{\sqrt{0^2 + 1}} + C \:=\:1\quad\Rightarrow\quad C = 1

    Therefore: . y(x) \;=\;\frac{x}{\sqrt{x^2+1}} + 1

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