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Math Help - Implicit Solutions

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    Super Member Aryth's Avatar
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    Implicit Solutions

    I have two problems that I'm having some serious difficulty with. It's probably because of my lack of knowledge of implicit solutions. If, while you solve these, you could explain a little more about implicit solutions I would be incredibly grateful.

    In these problems, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship does define y implicity as a function of x and use implicit differentiation.

    1.
    e^{xy} + y = x - 1,

    \frac{dy}{dx} = \frac{e^{-xy} - y}{e^{-xy} + x}


    2.
    \sin{(y)} + xy - x^3 = 2,

    y'' = \frac{6xy' + (y')^3\sin{(y)} - 2(y')^2}{3x^2 - y}
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    Quote Originally Posted by Aryth View Post
    I have two problems that I'm having some serious difficulty with. It's probably because of my lack of knowledge of implicit solutions. If, while you solve these, you could explain a little more about implicit solutions I would be incredibly grateful.

    In these problems, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship does define y implicity as a function of x and use implicit differentiation.

    1.
    e^{xy} + y = x - 1,

    \frac{dy}{dx} = \frac{e^{-xy} - y}{e^{-xy} + x}
    Implicit differentiation is just differentiating an arbitrary equation, instead of differentiating y=...

    What you can do in order to solve this problem is first differentiate the equation and then substitute \frac{dy}{dx} you're given and see if the equation is satisfied.

    e^{xy}+y=x-1
    Remember that y is a function of x.
    So the derivative of y wrt x will be \frac{dy}{dx} and not 0.
    In these problems, you'll mainly have to use the product rule, chain rule and quotient rule.

    Let's do it by steps.
    What if we differentiate e^{xy} with respect to x ?
    We know that its derivative is (xy)' e^{xy}
    Now what is the derivative of xy : (xy)' ?
    Remember that y is a function of x. It's not a constant.
    So you have to use the chain rule :
    (xy)'=(x)'y+x(y)'=1 \cdot y+x \cdot \frac{dy}{dx}=y+x \cdot \frac{dy}{dx}
    Hence differentiating e^{xy} will give :
    \left(y+x \cdot \frac{dy}{dx}\right) e^{xy}

    What if we differentiate y with respect to x ?
    That's simply \frac{dy}{dx}

    What if we differentiate x+1 with respect to x ?
    That's just 1.

    So if you differentiate your equation, you'll get :
    \left(y+x \cdot \frac{dy}{dx}\right) e^{xy}+\frac{dy}{dx}=1

    now substitute \frac{dy}{dx} by the formula you are given and see if the equation is satisfied. And you'll be done.
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    Super Member Aryth's Avatar
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    Awesome, got it. The problem I had was manipulating it to look exactly like the differential equation. Thanks for the help.
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