Question

Find the particular integral of
(x-1)^2 y" - 2(x-1)y' + 2y = (x-1)^2 y(0) = 3, y'(0) = 6

Answer 1

Given D.E is

$$(x-1)^2 y'' - 2(x-1)y' + 2y = (x-1)^2 \:,y(0) = 3, y'(0) = 6$$

Notice that solution of homogeneious equation $(x-1)^2 y'' - 2(x-1)y' + 2y =0$ is

$$y_1=(x-1)$$

We find second independent solution fron Liouville formula is

$$W_{y_1y_2}(x)=\begin{vmatrix} y_1&y_2\\y_1'&y_2' \end{vmatrix}=C_1e^{-\int \frac{-2(x-1)}{(x-1)^2}dx}=C_1(x-1)^2$$

Thus,

$$\dfrac{y_1y_2'-y_1'y_2}{y_1^2}=C_1\frac{(x-1)^2}{(x-1)^2}=C_1\implies (\frac{y_2}{y_1})'=C_1\\ \implies y_2=C_1(x-1)^2+C_2(x-1)$$

Hence, general solution of homogeneous equation is

$$y_o(x)=C_1(x-1)^2+C_2(x-1)$$

Now, using methood of variation of parameters and construct general solution of inhomogeneous equation ,

$y_1 ( x ) \:\text{and}\: y_ 2 ( x ) : y_ 0 ( x ) = C_ 1 y_ 1 ( x ) + C_ 2 y_ 2 ( x ) , where \: C_ 1 , C_ 2 \:\text{are arbitrary constants.}$

Thus,

$$\text{The derivatives of the unknown functions } C_ 1 ( x ) \& C_ 2 ( x )\: \\ \text{can be determined from the system of equations}$$

$$C^ ′_ 1 ( x ) y_ 1 ( x ) + C^ ′_ 2 ( x ) y_ 2 ( x ) = 0\\ C^ ′_ 1 ( x ) y^ ′_ 1 ( x ) + C^ ′_ 2 ( x ) y^ ′_ 2 ( x ) = f ( x )$$

Hence,

$${{C’_1}\left( x \right) = – \frac{{{y_2}\left( x \right)f\left( x \right)}}{{{W_{{y_1},{y_2}}}\left( x \right)}},\;\;}\kern-0.3pt {{C’_2}\left( x \right) = \frac{{{y_1}\left( x \right)f\left( x \right)}}{{{W_{{y_1},{y_2}}}\left( x \right)}}.}$$

Furthermore, knowing the derivatives of ${C’_1}\left( x \right)$ and ${C’_2}\left( x \right)$ ,one can find the $C_1(x)\& C_2(x)$ Thus,

$${{{C_1}\left( x \right) }={ – \int {\frac{{{y_2}\left( x \right)f\left( x \right)}}{{{W_{{y_1},{y_2}}}\left( x \right)}}dx} }+{ {A_1},\;\;}}\kern-0.3pt {{{C_2}\left( x \right) }={ \int {\frac{{{y_1}\left( x \right)f\left( x \right)}}{{{W_{{y_1},{y_2}}}\left( x \right)}}dx} }+{ {A_2}}}$$

where$A_1\&A_2$ are constants of integration.Then the general solution of the original nonhomogeneous equation will be expressed by the formula

$${y\left( x \right) }={ {C_1}\left( x \right){y_1}\left( x \right) + {C_2}\left( x \right){y_2}\left( x \right) } = {{\left[ { – \int {\frac{{{y_2}\left( x \right)f\left( x \right)}}{{{W_{{y_1},{y_2}}}\left( x \right)}}dx} + {A_1}} \right] \cdot}\kern0pt{ {y_1}\left( x \right) }} +\\ {{\left[ {\int {\frac{{{y_1}\left( x \right)f\left( x \right)}}{{{W_{{y_1},{y_2}}}\left( x \right)}}dx} + {A_2}} \right] \cdot}\kern0pt{ {y_2}\left( x \right) }} = {{{A_1}{y_1}\left( x \right) + {A_2}{y_2}\left( x \right) }+{ Y\left( x \right)}}$$

Where,

$${Y\left( x \right) } = {{y_2}\left( x \right)\int {\frac{{{y_1}\left( x \right)f\left( x \right)}}{{{W_{{y_1},{y_2}}}\left( x \right)}}dx} } – {{y_1}\left( x \right)\int {\frac{{{y_2}\left( x \right)f\left( x \right)}}{{{W_{{y_1},{y_2}}}\left( x \right)}}dx} }$$

Thus, from above fact by putting $f(x)=(x-1)^2$ and solving exactly as above we get the general solution is

$$C_1(x-1)-(x+\ln(x-1))(x-1)+C_2x(x-1)+x\ln(x-1)(x-1)$$

Thus particular solution is

$$- (x-1)(10x+\ln(x-1)-x\ln(x - 1)+3-\pi i + \pi xi)$$