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Show that $3y''+4xy'-8y=0$ has an integral which is a polynomial in x. Deduce the general solution.

In this problem, i tried in direct method. But i could not do. So, i did in following way

Since order is 2, i am assuming a polynomial of degree 2. Let $ y=ax^2+bx+c$ be the solution. By substitution, i get Solution as $y= (4/3)x^2+c$, c is some constant. Pls correct me if am wrong. Any other better way, pls suggest

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  • $\begingroup$ Your answer is correct only for $c=1$ so you made a slight error somewhere. I suspect the hint that there was a solution which was a polynomial in $x$ was a hint to try a small polynomial such as a first or second degree. I presume that you are currently learning how to solve second order linear equations given one solution? Any constant multiple of $4x^2+3$ will be a solution. $\endgroup$ – John Wayland Bales Aug 22 '18 at 21:51
  • $\begingroup$ Later you will learn how to solve such equations using power series. $\endgroup$ – John Wayland Bales Aug 22 '18 at 21:54
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If $y(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ is a solution with $a_n\neq 0$, then note that $3\,y''(x)$ is of degree $n-2$, but $4x\,y'(x)$ and $-8\,y(x)$ are of the same degree $n$. As $3\,y''(x)+4x\,y'(x)-8\,y(x)=0$, the leading terms of $4x\,y'(x)$ and $-8\,y(x)$ must cancel. This proves that $$4n\,a_n\,x^n-8\,a_n\,x^n=0\,,\text{ whence }n=2\,.$$ Thus, we may assume that $y(x)=x^2+bx+c$ is a solution. Now, $$0=3\,y''(x)+4x\,y'(x)-8\,y(x)=3\cdot 2+4x\,(2x+b)-8\,(x^2+bx+c)=-4b\,x+(6-8c)\,,$$ so $b=0$ and $c=\dfrac{3}{4}$. This means $$y(x)=x^2+\frac34$$ is a solution.

To find the general solution, we suppose that $y(x)=\left(x^2+\dfrac34\right)\,z(x)$ satisfies the differential equation. Plugging this in to get $$3\,\left(x^2+\frac34\right)\,z''(x)+\Biggl(12x+4x\,\left(x^2+\frac34\right)\Biggr)\,z'(x)=0\,.$$ In other words, $$z''(x)+\left(\frac{4}{3}\,x+\frac{4x}{x^2+\frac34}\right)\,z'(x)=0\,,$$ or $$\frac{\text{d}}{\text{d}x}\,\left(\left(x^2+\frac34\right)^2\,\exp\left(\frac{2}{3}\,x^2\right)\,z'(x)\right)=0\,.$$ Ergo, $$z'(x)=A'\,\left(\frac{\exp\left(-\frac{2}{3}\,x^2\right)}{\left(x^2+\frac{3}{4}\right)^2}\right)\text{ for some constant }A'\,.$$ In conclusion, $$y(x)=A'\,\left(x^2+\frac34\right)\,\int_0^x\,\frac{\exp\left(-\frac{2}{3}\,t^2\right)}{\left(t^2+\frac{3}{4}\right)^2}\,\text{d}t+B\,\left(x^2+\frac{3}{4}\right)$$ for some constant $B$. We may write $$y(x)=\small A\,\Biggl(\sqrt{\frac{2\pi}{3}}\,\left(x^2+\frac34\right)\,\text{erf}\left(\sqrt{\frac23}\,x\right)+x\,\exp\left(-\frac23\,x^2\right)\Biggr)+B\,\left(x^2+\frac{3}{4}\right)\,,$$ where $A:=\frac{2}{3}\,A'$ and $\text{erf}$ is the error function: $$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\,\int_0^x\,\exp\left(-t^2\right)\,\text{d}t\,.$$ We also have $A=\dfrac{1}{2}\,y'(0)$ and $B=\dfrac{4}{3}\,y(0)$. That is, $$y(x)=\small y(0)\,\left(\frac{4}{3}\,x^2+1\right)+y'(0)\,\Biggl(\frac{\sqrt{6\pi}}{8}\,\left(\frac43\,x^2+1\right)\,\text{erf}\left(\sqrt{\frac23}\,x\right)+\frac12\,x\,\exp\left(-\frac23\,x^2\right)\Biggr)\,.$$ Interestingly, we may write $$\left(\frac{\text{d}}{\text{d}x}+\frac{4x}{3}+\frac{2x}{x^2+\frac34}\right)\,\left(\frac{\text{d}}{\text{d}x}-\frac{2x}{x^2+\frac34}\right)=\frac{\text{d}^2}{\text{d}x^2}+\frac{4x}{3}\,\frac{\text{d}}{\text{d}x}-\frac{8}{3}\,.$$

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  • $\begingroup$ Sir i am amazed at the away u solved it... I am a beginner. Can you pls tell how did u write integrate directly t0 some $f(x)e^{g(x)}$ (in 2nd step after "in other words") $\endgroup$ – Magneto Aug 23 '18 at 4:41
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    $\begingroup$ I simply integrate $\displaystyle \frac{4x}{3}+\frac{4x}{x^2+\frac{3}{4}}$ to get $$\frac{2}{3}\,x^2+2\,\ln\left(x^2+\frac{3}{4}\right)\,.$$ Take the exponential of the fuction above to get $$\left(x^2+\frac34\right)^2\,\exp\left(\frac23\,x^2\right)\,.$$ This is the integral factor that you refer to. $\endgroup$ – Batominovski Aug 23 '18 at 15:13
  • $\begingroup$ Sir can u pls explain me how did u write 2 steps after "In conclusion". Actually i get $z = A' \int F(x) + B'$ --> how instead of indefinite integral, definite integral comes. And after this directly error function comes in. I am lost. Kindly elaborate. I am learning. $\endgroup$ – Magneto Aug 23 '18 at 15:42
  • $\begingroup$ Also, sir, can u pls see this --> math.stackexchange.com/questions/2891936/… $\endgroup$ – Magneto Aug 23 '18 at 15:42

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