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Answer:
Since ( y'' = 4 ), we integrate with respect to ( x ) to find( y' ):
[tex]y'' = 4 \implies y' = 4x + C_1[/tex]
Here, ( C_1 ) is the constant of integration.
2. Integrate again to find ( y ):
Now, integrate ( y' ) with respect to ( x ) to find ( y ):
[tex]{y' = 4x + C_1 \implies y = 2x^2 + C_1 x + C_2}[/tex]
Here, ( C_2) is another constant of integration.
3. Use the initial conditions to find the constants ( C_1 ) and ( C_2 ):
[tex]( y'(2) = -1 ):[/tex]
[tex]{y'(2) = 4(2) + C_1 = -1 \implies 8 + C_1 = -1 \implies C_1 = -9}
[/tex]
[tex]( y(2) = -1 ):[/tex]
[tex]{y(2) = 2(2)^2 + (-9)(2) + C_2 = -1 \implies 8 - 18 + C_2 = -1 \implies -10 + C_2 = -1 \implies C_2 = 9}[/tex]
[tex]y = 2x^2 - 9x + 9[/tex]
Thus, the solution to the initial value problem is:
[tex]y = 2x^2 - 9x + 9[/tex]