How do you solve second order initial value problems?
I don't understand how would solve these questions..for example
(d^2 y)/(dx^2 )=24x^2-10. When x=1, dy/dx = 3, and y = 5.
Can anyone explain how?
To solve the given second-order initial value problem, follow these steps:
1. Identify the given differential equation:
The given differential equation is (d^2 y)/(dx^2 ) = 24x^2-10.
2. Rewrite the equation:
In this step, we rewrite the second-order equation into two first-order equations. Let's introduce a new variable, u, such that u = dy/dx. Now, we have two first-order equations:
du/dx = 24x^2-10, (1) - (Equation 1)
dy/dx = u. - (Equation 2)
3. Solve Equation 1:
To solve Equation 1, integrate both sides with respect to x:
∫ du = ∫ (24x^2-10) dx
u = 8x^3-10x + C1, where C1 is the constant of integration.
4. Solve Equation 2:
To solve Equation 2, separate the variables and integrate:
∫ (1/u)du = ∫ dx
ln|u| = x + C2, where C2 is another constant of integration.
5. Solve for u:
By exponentiating both sides, we get:
|u| = e^(x+C2) = Ae^x, where A is a nonzero constant.
6. Determine the sign of A:
Since |u| = Ae^x, take into account that dy/dx = u. Therefore, the sign of A depends on the given initial condition for dy/dx.
7. Apply the given initial condition:
When x = 1, dy/dx = 3. Plug these values into Equation 2:
3 = u = Ae^1 = Ae
8. Find the value of A:
From the previous step, we have 3 = Ae. Solving for A, we have A = 3/e.
9. Determine the sign of A:
Since A = 3/e, and we know A is nonzero, we conclude that A is positive.
10. Revisit Equation 1:
We previously determined that u = 8x^3-10x + C1.
11. Substitute A into u:
Given that A = 3/e, we now have: u = (3/e)e^x = 3e^(x-1).
12. Revisit Equation 2:
Let's rewrite Equation 2 with the value of u we found:
dy/dx = 3e^(x-1).
13. Solve Equation 2:
To solve Equation 2, integrate both sides with respect to x:
∫dy = ∫3e^(x-1)dx.
14. Solve for y:
Integrating both sides of the equation, we get:
y = 3e^(x-1) + C3, where C3 is the constant of integration.
15. Apply the second initial condition:
When x = 1, y = 5. Plug these values into the equation:
5 = 3e^(1-1) + C3,
5 = 3 + C3,
C3 = 5 - 3 = 2.
16. Final solution:
Substitute C3 into the previous equation to obtain:
y = 3e^(x-1) + 2.
So, the solution to the given second-order initial value problem is y = 3e^(x-1) + 2.