Given the fundamental system y_1=x , y_2=x^2 , y_3=x^3 of a…

Given the fundamental system $y_1=x$, $y_2=x^2$, $y_3=x^3$ of a homogeneous linear ODE, find the particular solution satisfying $y(1)=0$, $y'(1)=-1$, $y''(1)=2$.

Step 1: General solution

Every solution is a linear combination of the fundamental system: $y=C_1x+C_2x^2+C_3x^3.$

Its derivatives are

\[ y'=C_1+2C_2x+3C_3x^2,\qquad y''=2C_2+6C_3x. \]

Step 2: Apply the initial conditions at $x=1$

\[ \begin{aligned} y(1)&=C_1+C_2+C_3=0,&&(1)\\ y'(1)&=C_1+2C_2+3C_3=-1,&&(2)\\ y''(1)&=2C_2+6C_3=2.&&(3) \end{aligned} \]

Step 3: Solve the linear system

Divide (3) by 2: \[C_2+3C_3=1. \tag{3'}\]

Subtract (1) from (2): \[C_2+2C_3=-1. \tag{4}\]

Subtract (4) from (3'): $C_3=2.$

Back-substitute into (4): $C_2+4=-1\Rightarrow C_2=-5$. Back-substitute into (1): $C_1-5+2=0\Rightarrow C_1=3$.

\[ C_1=3,\qquad C_2=-5,\qquad C_3=2. \]

Step 4: Particular solution

The solution is recorded in the result below.

Step 5: Verification

\[ y(1)=3-5+2=0.\ \checkmark \]

\[ y'(x)=3-10x+6x^2\ \Longrightarrow\ y'(1)=3-10+6=-1.\ \checkmark \]

\[ y''(x)=-10+12x\ \Longrightarrow\ y''(1)=-10+12=2.\ \checkmark \]