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Thread: Euler equation having double roots or repeated roots as a solution

  1. #1
    Senior Member Vinod's Avatar
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    Euler equation having double roots or repeated roots as a solution

    Hello,
    If the Euler equation have a double roots as a solution,second solution will be $y_2=x^r\ln{x}$.


    What is its proof? or how it can be derived? To find a second solution,we will use the fact that constant times a solution is also a solution to linear homogeneous differential equations. Now why do we choose $\ln{x}$ as a constant?why not any other constant?
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  2. #2
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    Re: Euler equation having double roots or repeated roots as a solution

    "Euler's equation" (also called an "Euler type equation" or "equipotential equation") is a differential equation of the form $\displaystyle a_nx^n\frac{d^ny}{dx^n}+ a_{n-1}x^{n-1}\frac{d^{n-1}y}{dx^{n-1}}+ \cdot\cdot\cdot+ a_1x\frac{dy}{dx}+ a_0y= 0$. Since we are talking about a double root, it is sufficient to assume second order: $\displaystyle a_2x^2\frac{d^2y}{dx^2}+ a_1x\frac{dy}{dx}+ a_0y= 0$.

    Using the substitution $\displaystyle v= ln(x)$, Euler's equation reduces to an equation with constant coefficients:
    $\displaystyle \frac{dy}{dx}= \frac{dy}{dv}\frac{dv}{dx}= \frac{1}{x}\frac{dy}{dv}$ and $\displaystyle \frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dv}\right)= -\frac{1}{x^2}\frac{dy}{dv}+ \frac{1}{x^2}\frac{d^2y}{dv^2}$ so the equation becomes $\displaystyle a_2x^2\left(\frac{1}{x^2}\frac{d^2y}{dv^2}- \frac{1}{x^2}\frac{dy}{dv}\right)+ a_1x\left(\frac{1}{x}\frac{dy}{dv}\right)+ a_0y= a_2\frac{d^2y}{dv^2}+ (a_1- a_2)\frac{dy}{dv}+ a_0y= 0$.

    That "constant coefficients" differential equation has characteristic equation $\displaystyle a_2r^2+ (a_1- a_2)r+ a_0= 0$. If it has a double root, say $\displaystyle r= r_0$ then the differential equation has solutions of the form $\displaystyle y= e^{r_0v}$ and $\displaystyle y= ve^{r_0v}$. Since $\displaystyle v= ln(x)$, those solutions are $\displaystyle y= e^{r_0 ln(x)}= e^{ln(x^{r_0})}= x^{r_0}$ and $\displaystyle y= ln(x)e^{r_0ln(x)}= ln(x)e^{ln(x^{r_0})}= x^{r_0}ln(x)$.
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  3. #3
    Member Walagaster's Avatar
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    Re: Euler equation having double roots or repeated roots as a solution

    Here's another way to see it. Let's call the operator $L(y) = a_2x^2y'' + a_1xy' + a_0y$. If you try for a solution of the form $y = x^r$ you get$$
    L(x^r)=a_2r(r-1)x^r +a_1rx^r +a_0x^r = p(r)x^r$$We could call $p(r) = a_2r(r-1)+a_1r +a_0$ the characteristic polynomial of the DE. Now, if $r$ is a root of $p(r)$ so $p(r)=0$ then the above equation gives $L(x^r) = p(r)x^r= 0$, so $x^r$ is a solution of the DE. Now let's suppose $r$ is a double root of $p(r)$, so both $p(r)=0$ and $p'(r) = 0$. Now let's take the result $L(x^r) = p(r)x^r$ and differentiate both sides with respect to r, not $x$. Assuming we can switch the order of differentiation we get$$
    \frac \partial {\partial r}L(x^r) = L(\frac\partial {\partial r}x^r) = \frac\partial {\partial r}(p(r)x^r)$$Now remember that when we differentiate $x^r$ with respect to $r$ we are differentiating an exponential function to a base other than $e$ because the variable $r$ is in the exponent. So carrying out the derivation in the above line gives$$
    L(x^r\ln x) = p'(r)x^r + p(r)x^r\ln x = 0$$because both $p(r)=0$ and $p'(r) = 0$. This tells you that $x^r\ln x$ is a second solution.
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    Re: Euler equation having double roots or repeated roots as a solution

    Also, by the method of Reduction of Order:

    Given that $r$ is a double root of the polynomial $p(x)=a_2x^2+ (a_1- a_2)x+ a_0 = (x-r)^2$, we know that $p(r) = 0$ and also $p'(r) = 0$ so that \begin{align}2a_2r + a_1 - a_2 &= 0 \\ 2a_2r + a_1&=a_2\end{align}

    Also given that $y_1=x^r$ is a solution of $a_2x^2y''+a_1xy'+a_0y=0$, we seek a second solution $y_2(x)=v(x)y_1(x)$. That is \begin{align}y_2&=vx^r \\ \text{Thus} \; y_2' &= v'x^r + rvx^{r-1} \\ \text{and} \; y_2''&=v''x^r+2rv'x^{r-1}+r(r-1)vx^{r-2}\end{align}

    Substituting $y_2$ and it's derivatives in the original equation we get \begin{align}a_2x^2y_2''+a_1xy_2'+a_0y_2 &= 0 \\ a_2x^2\big(v''x^r+2rv'x^{r-1}+r(r-1)vx^{r-2}\big) + a_1x\big(v'x^r + rvx^{r-1}\big) + a_0vx^r &= 0 \\ a_2v''x^{r+2}+(2a_2r+a_1)v'x^{r+1}+\big(a_2x^2 r(r-1)x^{r-2} + a_1xrx^{r-1} + a_0x^r\big)v &= 0 \end{align}
    Now the coefficient of the term in $v$ is equal to zero because $x^r$ is a solution of the original equation, and the parenthesis in the second term is equal to $a_2$ by the first work from above so we have \begin{align}\big(a_2xv''+a_2v'\big)x^{r+1} &= 0 \\ xw'+w &= 0 \\ \frac{\mathrm d}{\mathrm dx}(xw) &= 0 \\ xw &= c_1 \\ v' = w &= \frac{c_1}{x} \\ v &= c_1\ln{(x)}+c_2 \end{align}
    And so $$y_2 = vy_1 = (c_1\ln{(x)}+c_2)x^r$$

    The general solution of the original equation is thus \begin{align}y &= c_3y_1 + c_4y_2 \\ &= c_3x^r + c_4(c_1\ln{(x)}+c_2)x^r \\ &= c_5x^r + c_6x^r\ln{(x)}\end{align}
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  5. #5
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    Re: Euler equation having double roots or repeated roots as a solution

    I forgot to say in the line $xw'+w=0$that $w=v'$.
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