Question

Solve this Differentiate equation –

$\\ \bf {x}^{2} \dfrac{ {d}^{2}y}{ {dx}^{2} } + 3x \dfrac{dy}{dx} + y = \dfrac{1}{ {(1 - x)}^{2} }$​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

Solve this Differentiate equation –

$\sf{\displaystyle\mathrm{x}^2\:\frac{\mathrm{d}^2\:\mathrm{y}}{\mathrm{dx}^2}+3\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=\frac{1}{\left(1-\mathrm{x}\right)^2}}$

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♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{y=\frac{c_1}{x}+\frac{c_2\ln \left(x\right)}{x}+\frac{\ln \left(x\right)-\ln \left(x-1\right)}{x}}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\sf{\displaystyle x^2\frac{d^2y}{dx^2}+3x\frac{dy}{dx}+y=\frac{1}{\left(1-x\right)^2}}$

Second order Euler non-homogeneous differential equation

A second order Euler non-homogeneous ODE has the form of  ax²y''+ bxy' + cy = gx

$\sf{\mathrm{Substitute\quad }\dfrac{d^2y}{dx^2},\:\dfrac{dy}{dx}\mathrm{\:with\:}y''\:,\:y'\:}$

$\sf{x^2y''\:+3xy'\:+y=\dfrac{1}{\left(1-x\right)^2}}$

General solution for a(x)y''+ b(x)y'+ c(x)y = g(x)

$\mathrm{The\:general\:solution\:to\:}a\left(x\right)y''+b\left(x\right)y'+c\left(x\right)y=g\left(x\right)\mathrm{\:can\:be\:written\:as}$

$y=y_h+y_p$

$y_h\mathrm{\:is\:the\:solution\:to\:the\:homogeneous\:ODE\:}a\left(x\right)y''+b\left(x\right)y'+c\left(x\right)y=0$

$y_p\mathrm{,\:the\:particular\:solution,\:is\:any\:function\:that\:satisfies\:the\:non-homogeneous\:equation}\:$

$\bigstar\:\sf{Find \: y_{h} by solving x^{2} y^{\prime \prime}+3 x y^{\prime}+y=0 }$

Second order Euler homogeneous differential equation

A second order Euler homogeneous ODE has the form of  ax²y''+ bxy' + cy = 0

$\text { For an equation } a x^{2} y^{\prime \prime}+b x y^{\prime}+c y=0, \text { assume a solution of the form } x^{r}$

$\mathrm{Rewrite\:the\:equation\:with\:}y=x^r$

$x^2\left(\left(x^r\right)\right)''\:+3x\left(\left(x^r\right)\right)'\:+x^r=0$

$\implies x^r\left(r^2+2r+1\right)=0$

$\text { For one real root } r \text { , the general solution takes the form: } y=c_{1} x^{r}+c_{2} \ln (x) x^{r}$

$c_1x^{-1}+c_2\ln \left(x\right)x^{-1}$

$y=\dfrac{c_1}{x}+\dfrac{c_2\ln \left(x\right)}{x}$

$\bigstar\:\text { Find } y_{p} \text { that satisfies } x^{2} y^{\prime \prime}+3 x y^{\prime}+y=\dfrac{1}{(1-x)^{2}}$

$x^2y''\:+3xy'\:+y=\dfrac{1}{\left(1-x\right)^2}$

$\mathrm{Divide\:both\:sides\:by\:}x^2$

$\displaystyle y''\:+\frac{3y'\:}{x}+\frac{y}{x^2}=\frac{1}{x^2\left(1-x\right)^2}$

$\text{Where y_{1} and y_{2} are solutions of homogeneous equation}\\ y_{h}=c_{1} y_{1}+c_{2} y_{2} u_{1} and u_{2} are solutions to the system of equations:$

$\left(\begin{array}{c}\prime_{1}^{\prime} y_{1}+u_{2}^{\prime} y_{2}=0 \\\\u_{1}^{\prime}+u_{2} y_{2}^{\prime}=g(x)\end{array}\right)$

Which implies :

$u_1=\int \:-\dfrac{y_2g\left(x\right)}{W\left(y_1,\:y_2\right)}dx$

$u_2=\int \dfrac{y_1g\left(x\right)}{W\left(y_1,\:y_2\right)}dx$

$\mathrm{Where\:Wronskian}\:W\left(y_1,\:y_2\right)=y_1y_2^{'\:}-y_1^{'\:}y_2$

Homogeneous solutions :

$y_1=\dfrac{1}{x}$

$y_2=\dfrac{\ln \left(x\right)}{x}$

$\begin{array}{l}y_{1}^{\prime}: \quad-\dfrac{1}{x^{2}} \\\\y_{2}: \dfrac{1-\ln (x)}{x^{2}}\end{array}$

$W\left(y_1,\:y_2\right)=y_1y_2^{'\:}-y_1^{'\:}y_2$

$\implies\displaystyle W\left(y_1,\:y_2\right)=\frac{1}{x}\cdot \frac{1-\ln \left(x\right)}{x^2}-\left(-\frac{1}{x^2}\right)\frac{\ln \left(x\right)}{x}$

$\implies W\left(y_1,\:y_2\right)=\dfrac{1}{x^3}$

$\displaystyle u_1=\int \:-\frac{y_2g\left(x\right)}{W\left(y_1,\:y_2\right)}dx$

$\implies\displaystyle u_1=\int \:-\frac{\frac{\ln \left(x\right)}{x}\cdot \frac{1}{x^2\left(1-x\right)^2}}{\frac{1}{x^3}}dx$

$\implies u_1=-\dfrac{\ln \left(x\right)}{1-x}+\ln \left(x\right)-\ln \left(x-1\right)$

$u_2=\int \dfrac{y_1g\left(x\right)}{W\left(y_1,\:y_2\right)}dx$

$\implies u_2=\int \dfrac{\frac{1}{x}\cdot \dfrac{1}{x^2\left(1-x\right)^2}}{\dfrac{1}{x^3}}dx$

$\implies\:u_2=\dfrac{1}{1-x}$

$\bf{y_p=u_1y_1+u_2y_2}$

$\displaystyle y_p=\left(-\frac{\ln \left(x\right)}{1-x}+\ln \left(x\right)-\ln \left(x-1\right)\right)\frac{1}{x}+\frac{1}{1-x}\cdot \frac{\ln \left(x\right)}{x}$

$\implies\displaystyle\:y_p=\frac{\ln \left(x\right)-\ln \left(x-1\right)}{x}$

$\sf{\text { A particular solution } y_{p} \text { to } x^{2} y^{\prime \prime}+3 x y^{\prime}+y=\dfrac{1}{(1-x)^{2}} \text { is: }}$

$y_p=\dfrac{\ln \left(x\right)-\ln \left(x-1\right)}{x}$

$\mathrm{The\:general\:solution\:}y=y_h+y_p\mathrm{\:is:}$

$\boxed{\sf{y=\frac{c_1}{x}+\frac{c_2\ln \left(x\right)}{x}+\frac{\ln \left(x\right)-\ln \left(x-1\right)}{x}}}$

Refer to attachment for plotting

Answer 2

$\bf\huge{\underline{\pink{ ♣ Given }}}$

$\frac{x²d²y}{dx²} + 3 \times \frac{dy}{dx} + y = \frac{1}{(l - x)²} \\Let x = et$

$➡log \: x = t \: and \: we \: take \frac{d}{d} D\\➡[D(D-1+3D+1] y = (\frac{1}{1}-x)² \\➡(D + 1)²y = \frac{1}{(1 - x)²} ➡ (1) \\ The \: homogeneous \: solution \\ will \: be \: (0 + 1)²$

$C.F = (c, + \: c² \: t) {e}^{ - t} \\ [{e}^{ - t} = ({e}^{ t})¹ = {x}^{ - 1} ]$

$Now, \\ P. I = \frac{1}{(D + 1)} \frac{1}{(D + 1)} {(1 - x)}^{ - 2} \\ = \frac{1}{(D + 1)} {x}^{ - 1} {(1 - x)}^{ - 1} \\ = \frac{1}{(D + 1)} \: {x}^{ - 1} {(1 - x)}^{ - 2}dx \\ = {x}^{ - 1} {x}^{ - 1} {(1 - x)}^{ - 1} dx$

$= {x}^{ - 1} \frac{dx}{x(1 - x)} \\ = {x}^{ - 1}[ \frac{1}{x} + \frac{1}{1 - x} ]dm \\ = {x}^{ - 1}[log \: x - log(1 - x)] \\ = {x}^{ - 1} \: log [ \frac{x}{(1 - x)} ]$

The solution is

$y = ( {c}^{1} + {c}^{2} \: log \: x) {x}^{ - 1} + {x}^{ - 1} log [x/(1 - x)]$

Formula:-

$simple \: way \: to \: calulate \: P. I. \\ here \: in \: this \: case \\ y = \frac{1}{(D + 1)}y\frac{1}{(D + 1)}y\frac{1}{(x - 2)}² \\ Next \: step \: is \: to \: conuert \: that \\ ( \frac{1}{D + y} ) \: to \: {x}^{ - 1} \: as \: we \: got \: in \: CF \\ :e, P. I. = \frac{1}{(D +D )} {x}^{ - 1}$

$\frac{1}{(x - 2)} ² \frac{1}{x} - 1 \: dx \\ then \: we \: get \: to \: simply \: it \: and \\ again \: convert \: this \: \frac{1}{(D + 1)} \: to \: \\ {x}^{ - 1}$

$\bf{\underline{\green{ Hope \: it \: helps \: you }}}$