Respuesta :
Answer:
Therefore the particular solution of the given differential equation is
[tex]Y(x)=\frac{1}{6} x^2 ln x[/tex]
Step-by-step explanation:
The given ordinary differential equation is
[tex]x^2y"-3xy' +4y=11x^2[/tex]
If y₁(x) =x² is a solution of ODE then it will be satisfy the ODE
y₁'(x)= 2x and y₁"(x)=2
Putting the value of y₁'(x) and y₁"(x) in x²y"-3xy'+4y=0 we get
x².2-3x.2x+4x²= 2x²-6x²+4x²=0
Therefore y₁(x) is a solution of the given differential equation.
Again,
y₂(x) =(x²ln x ) is a solution of ODE then it will be satisfy the ODE.
[tex]y'_2=2x ln x+ x^2\times\frac{1}{x}[/tex]
[tex]= 2x ln x+x[/tex]
[tex]y"_2(x)=2x\times \frac{1}{x} +2lnx+1[/tex] = 3+2 ln x
Putting the value of y₂'(x) and y₂"(x) in x²y"-3xy'+4y=0 we get
[tex]x^2 (3+2 ln x)-3x( 2x ln \ x+x)+4x^2\ ln\ x[/tex] =0
Therefore y₂(x) is a solution of the given differential equation.
The wronskian of y₁(x) and y₂(x) is
[tex]W(y_1,y_2)(x)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|[/tex]
[tex]=\left|\begin{array}{cc}x^2&x^2 ln x\\2x&2x ln x+x\end{array}\right|[/tex]
=x²(2x ln x+x)-x²ln x(2x)
=2x³ ln x +x³ - 2x³ln x
=x³≠0
Here [tex]g(x)=\frac{y_2(x)}{y_2(x)} = ln x[/tex]
The particular solution is
[tex]Y(x)=- y_1(x)\int\frac{y_2(x).g(x)}{W(y_1,y_2)(x)}dx + y_2(x)\int\frac{y_1(x).g(x)}{W(y_1,y_2)(x)}dx[/tex]
[tex]=-x^2\int\frac{x^2 ln x . ln x}{x^3} dx+x^2ln x\int \frac{x^2ln x}{x^3} dx[/tex]
[tex]=-x^2\int \frac{(lnx)^2}{x} dx+x^2 ln x\int \frac{ln x}{x} dx[/tex]
[tex]=-x^2\int u^2 du+x^2ln x\int u dx[/tex] Let ln x =u [tex]\Rightarrow \frac{1}{x} dx=du[/tex]
[tex]=-x^2 \frac{u^3}{3} +x^2 ln x \frac{u^2}{2}[/tex]
[tex]=-\frac{x^2(lnx)^3}{3} +\frac{x^2(lnx)^3}{2}[/tex]
[tex]=\frac{1}{6}x^2 (ln x)^3[/tex]
Therefore the particular solution of the given differential equation is
[tex]Y(x)=\frac{1}{6} x^2 ln x[/tex]
