Respuesta :
Here is the right and correct question:
The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2,
[tex]y_2 = y_1 (x) \int\limits \dfrac{e ^{-\int\limits P(x) dx} }{y^2_1 (x)} dx \ \ \ \ \ (5)[/tex]
as instructed, to find a second solution [tex]y_2(x)[/tex]
[tex](1-2x-x^2)y''+2(1+x)y' -2y =0; \ \ \ y_1=x+1[/tex]
Answer:
[tex]y_2 = -2-x^2-x[/tex]
Step-by-step explanation:
Let take a look at the differential equation:
[tex](1-2x-x^2)y''+2(1+x)y' -2y =0[/tex]
So; [tex]y''+ \dfrac{2(1+x)}{(1-2x-x^2)}y' - \dfrac{2}{(1-2x-x^2)}y =0[/tex]
where;
[tex]P(x) = \dfrac{2(1+x)}{(1-2x-x^2)}[/tex] ;
Also:
[tex]Q(x) = \dfrac{-2}{(1-2x-x^2)}[/tex]
The task is to find the value of [tex]y_2(x)[/tex] by using the reduction formula [tex]y_2 = y_1 (x) \int\limits \dfrac{e^{-\int\limits P(x) dx }}{y_1^2(x)}dx[/tex] such that [tex]y_1(x) =x+1[/tex]
simplifying [tex]y_2 = y_1 (x) \int\limits \dfrac{e^{-\int\limits P(x) dx }}{y_1^2(x)}dx[/tex] ;we have:
[tex]y_2 =(x+1) \int\limits \dfrac{e ^{-\int\limits \frac{2(1+x)}{(1-2x-x^2)}}}{(x+1)^2} \ \ dx[/tex]
[tex]y_2 =(x+1) \int\limits \dfrac{e ^{\int\limits \frac{-2(1+x)}{(1-2x-x^2)}}}{(x+1)^2} \ \ dx[/tex]
[tex]y_2 =(x+1) \int\limits \dfrac{e^{In(1-2x-x^2)}}{(x+1)^2}\ \ dx[/tex]
[tex]y_2 =(x+1) \int\limits \dfrac{(1-2x-x^2)}{(x+1)^2} \ \ dx[/tex]
[tex]y_2 =(x+1) \int\limits \dfrac{1}{(x+1)^2}-\dfrac{2x}{(x+1)^2}- \dfrac{x^2}{(x+1)^2} \ \ dx[/tex]
Let assume that [tex]I_1[/tex] = [tex]\int\limits \dfrac{-2x}{(x+1)^2} \ \ dx[/tex]
[tex]= \int\limits \dfrac{-(2x+2-2) }{(x+1)^2} \ \ dx[/tex]
[tex]= \int\limits \dfrac{-(2x+2) }{(x+1)^2} + \dfrac{2}{(x+1)^2} \ \ dx[/tex]
[tex]=- In(x+1)^2 - \dfrac{2}{(x+1)}[/tex]
Also : Let [tex]I_2 = \int\limits \dfrac{x^2}{(x+1)^2} \ \ dx[/tex]
[tex]= \int\limits \dfrac{(x+1-1)^2}{(x+1)^2} \ \ dx[/tex]
[tex]= \int\limits \dfrac{(x+1)^2+1-2(x+1)}{(x+1)^2} \ \ dx[/tex]
[tex]= \int\limits \ 1 + \dfrac{1}{(x+1)^2}- \dfrac{2}{(x+1)} \ \ dx[/tex]
[tex]= x - \dfrac{1}{(x+1)}- 2 \ In (x+1)[/tex]
Replacing the value of [tex]I_1[/tex] and [tex]I_2[/tex] in the equation
[tex]y_2 =(x+1) \int\limits \dfrac{1}{(x+1)^2}-\dfrac{2x}{(x+1)^2}- \dfrac{x^2}{(x+1)^2} \ \ dx[/tex]
[tex]y_2 =(x+1) [ \ \int\limits \dfrac{-1}{(x+1)}+ (-In(x+1)^2-\dfrac{2}{(x+1)})-(x-\dfrac{1}{(x+1)}-2 In(x+1))][/tex]
[tex]y_2 =(x+1) [ \ \int\limits \dfrac{-1}{(x+1)}- In(x+1)^2-\dfrac{2}{(x+1)}-x+\dfrac{1}{(x+1)}+2 In(x+1))][/tex]
[tex]y_2 =(x+1) [ \ \int\limits \dfrac{-1}{(x+1)}- 2In(x+1) -\dfrac{2}{(x+1)}-x + \dfrac{1}{(x+1)} +2 In(x+1)][/tex]
[tex]y_2 = -2-x(x+1)[/tex]
Therefore;
[tex]y_2 = -2-x^2-x[/tex]
