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Math Help - calculus - homogenous de

  1. #1
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    Exclamation calculus - homogenous de

    find the gen. solution of dy/dx + ylogy = ye^x
    using the substituion u = logy

    thankyou!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    Let's do the substitution : u=\ln y \Leftrightarrow y=\exp u\implies \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u}  {\mathrm{d}x}\exp u

    Substitute y=\exp u and \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u}  {\mathrm{d}x}\exp u in the equation : \frac{\mathrm{d}y}{\mathrm{d}x}+y\ln y=y\exp x \Leftrightarrow \frac{\mathrm{d}u}{\mathrm{d}x}\exp u+u \exp u = \exp u \exp x

    Simplify by \exp u and the equation becomes  \frac{\mathrm{d}u}{\mathrm{d}x}+u =\exp x

    This is a first order ODE hence you've to solve the homogeneous equation first : \frac{\mathrm{d}u}{\mathrm{d}x}+u =0. It'll give a solution u_H. Then, you need to find a particular solution u_P of E. As the second member is \exp x, you may try u_P(x)=\lambda \exp x. (find \lambda by replacing u_P by its expression in E) The solutions of E will be given by the sum u=u_P+u_H and then you can use y=\ln u to get the solutions in terms of y.
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  3. #3
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    thanks for your help! ok so i understand up to du/dx + u = e^x, but then i'm confused as to how to continue solving?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    The first thing to do is to solve the homogeneous equation : \frac{\mathrm{d}u}{\mathrm{d}x}+u=0 \Leftrightarrow\frac{\mathrm{d}u}{\mathrm{d}x}=-u \implies the solutions are u_H(x)=K\exp(- x)

    Then, we have to find a particular solution u_P of \frac{\mathrm{d}u}{\mathrm{d}x}+u=\exp x. Using my own advice, I try with u_P(x)=\lambda \exp x \implies \frac{\mathrm{d}u_P}{\mathrm{d}x}=\lambda\exp x. Let's replace this in the equation and solve for \lambda :

    \frac{\mathrm{d}u_P}{\mathrm{d}x}+u_P=\exp x \implies \lambda\exp x+\lambda\exp x=\exp x

    Simplifying both sides by \exp x, we get \lambda=\frac{1}{2} whence u_P(x)=\frac{1}{2}\exp x

    The solutions of the ODE in terms of u are given by \boxed{u(x)=u_H(x)+u_P(x)=K\exp(-x)+\frac{1}{2}\exp x}

    Using y=\exp u, we get the solutions of \frac{\mathrm{d}y}{\mathrm{d}x} + y\ln y = y\exp x : \boxed{y(x)=\exp\left(K\exp(-x)+\frac{1}{2}\exp x\right)}
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  5. #5
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    Quote Originally Posted by mathshelpbook View Post
    find the gen. solution of dy/dx + ylogy = ye^x
    using the substituion u = logy

    thankyou!
    Also asked by a different member here: http://www.mathhelpforum.com/math-he...tion-help.html

    and replied too.
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