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Solve the initial value problem '' = 4 ; 2
( )=− 1, ' 2 ( )=− 1

Sagot :

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Answer:

1. Integrate the differential equation:

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 ):

First, use the initial condition

[tex]( y'(2) = -1 ):[/tex]

[tex]{y'(2) = 4(2) + C_1 = -1 \implies 8 + C_1 = -1 \implies C_1 = -9}

[/tex]

Next, use the initial condition

[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]

4. Write the solution:

[tex]y = 2x^2 - 9x + 9[/tex]

Thus, the solution to the initial value problem is:

[tex]y = 2x^2 - 9x + 9[/tex]

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