Answer:
y = -2*x^3 - x + 2
Step-by-step explanation:
We want to solve the differential equation:
y'' + 12*x = 0
such that:
y(0) = 2
y'(0) = -1
We can rewrite our equation to:
y'' = -12x
if we integrate at both sides, we get:
[tex]\int {y''} \, dx = y'= \int {-12x} \, dx[/tex]
Solving that integral we can find the value of y', so we will get:
y' = -12* (1/2)*x^2 + C = -6*x^2 + C
where C is the constant of integration.
Evaluating y' in x = 0 we get:
y'(0) = -6*0^2 + C = C
and for the initial value problem, we know that:
y'(0) = -1
then:
y'(0) = -1 = C
C = -1
So we have the equation:
y' = -6*x^2 - 1
Now we can integrate again, to get:
y = -6*(1/3)*x^3 - 1*x + K
y = -2*x^3 - x + K
Where K is the constant of integration.
Evaluating or function in x = 0 we get:
y(0) = -2*0^3 - 0 + K
y(0) = K
And by the initial value, we know that: y(0) = 2
Then:
y(0) = 2 = K
K = 2
The function is:
y = -2*x^3 - x + 2
