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How to find the exact solution for the following differential equation? $$\begin{cases}\dfrac{d^2y}{dx^2}+3x\dfrac{dy}{dx}-6y=0 \\ \\y(0)=1\\y'(1)=0.1\\0<x<1 \end{cases}$$

I don't know how to solve this question. I was trying to find complementary function but it has non constant coefficient. Then I tried to solve it using power series but couldn't solved.

Any help is appreciated.

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  • $\begingroup$ Forgetting the conditions, the general solution write $$y=c_1 \left(3 x^2+1\right)+c_2 \,e^{-\frac{3 x^2}{2}} H_{-3}\left(\sqrt{\frac{3}{2}} x\right)$$ where appears the Hermite polynomial. The nightmare starts when introducing the conditions. $\endgroup$ – Claude Leibovici Jun 10 '18 at 12:10
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Using the power series method

$$ y = \sum_{n=0}^\infty c_nx^n, \quad xy' = \sum_{n=0}^\infty nc_n x^n $$ $$ y'' = \sum_{n=2} n(n-1)c_nx^n = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2}x^n $$

Plugging these into the equation, we find a recurrence relation

$$ (n+2)(n+1)c_{n+2} + 3(n-2)c_n = 0 $$

Clearly, the odd and even coefficients are separated from each other. The even series is finite, with

$$ c_2 = 3c_0, c_{n>2} = 0 $$

Therefore, we can write $$ y(x) = c_0 (1+3x^2) + c_1 y_2(x) $$

where $y_2(x)$ is the remaining odd power series with $y_2(0) = 0$ and ${y_2}'(0)=1$

We can derive a closed form expression for the coefficients, but a better method would be to use reduction of order. Substitute $y_2(x) = (1+3x^2)u(x)$ to get a new equation

$$ u'' + \left(3x + \frac{12x}{1+3x^2} \right)u' = 0 $$

where

$$ u(0) = 0, \quad u'(0) = 1 $$

Integrating and applying the above conditions, we find

$$ u'(x) = \frac{e^{-3x^2/2}}{(1+3x^2)^2}, \quad u(x) = \int_0^x \frac{e^{-3t^2/2}}{(1+3t^2)^2} dt $$

The last step is to determine the free constants. We apply the given B.C.s

\begin{align} y(0) &= c_0 = 1 \\ y'(1) &= 6 + c_1 {y_2}'(1) = 0.1 \end{align}

where numerical integration must be performed to find

$$ {y_2}'(1) = 4u'(1) + 6u(1) = \frac{e^{-3/2}}{4} + 6\int_0^1\frac{e^{-3t^2/2}}{(1+3t^2)^2} dt \approx 2.213185 $$

$$ \implies c_1 = -\frac{5.9}{{y_2}'(1)} \approx -2.665481 $$

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  • $\begingroup$ "$y_2'(0)=1$ " How this came ? I am not getting this $\endgroup$ – Prashant More Jun 10 '18 at 9:12
  • $\begingroup$ The second solution came from factoring the leading constant out of the odd series $$y_2 = x + \frac{c_3}{c_1} x^3 + \frac{c_5}{c_1}x^5 + \dots$$ Differentiate the above. $\endgroup$ – Dylan Jun 10 '18 at 9:20
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One solution of the D.E. (not satisfying your boundary condition) is $y = 1 + 3 x^2$. Another can then be found using reduction of order.

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  • $\begingroup$ +1 Robert $K(3x^2+1)$ is a solution $\endgroup$ – Isham Jun 8 '18 at 19:12
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    $\begingroup$ Yes, but you want two linearly independent solutions, then use the boundary condition to find the coefficients. $\endgroup$ – Robert Israel Jun 8 '18 at 19:18
  • $\begingroup$ How to find that ? $\endgroup$ – Prashant More Jun 8 '18 at 19:51
  • $\begingroup$ How to find the $1+3x^2$? Look at what would be needed for a polynomial solution. $\endgroup$ – Robert Israel Jun 8 '18 at 22:55
  • $\begingroup$ so i tried solving it on my own using reduction of order I took $y_1=1+3x^2$ and $y_2=y_1u$ and $u$ is function of $x$ and I got $$u=c_1\int\frac{e^\frac{-3x^2}{2}}{(1+3x^2)^2}+c_2$$ but I don't know how to solve this type of integral, Is I am going in right way? If you can solve further plz post it . $\endgroup$ – Prashant More Jun 9 '18 at 7:22

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