let d2y/dx2= 6x. Find a solution to the differential equation that is continuous for negative infinity to positive infinity and whose graph passes through the point (0,1) and has a horizontal tangent there.
To find a solution to the given differential equation, we need to integrate it twice.
Step 1: Find the first antiderivative:
We integrate both sides of the equation with respect to x:
∫d^2y/dx^2 dx = ∫6x dx
This becomes:
d/dx(dy/dx) = 3x^2 + C1
Integrating one more time:
dy/dx = ∫(3x^2 + C1) dx
Integrating the right side:
dy/dx = x^3 + C1x + C2
Step 2: Solve for y:
We have the first derivative of y, so we need to find y. To do this, we integrate the expression we obtained above with respect to x:
∫dy/dx dx = ∫(x^3 + C1x + C2) dx
This simplifies to:
y = ∫(x^3 + C1x + C2) dx
Integrating each term:
y = (1/4)x^4 + (C1/2)x^2 + C2x + C3
Step 3: Use the initial condition to find the particular solution:
We know that the graph passes through the point (0,1) and has a horizontal tangent there. Substituting x = 0 and y = 1 into the equation, we get:
1 = (1/4)(0)^4 + (C1/2)(0)^2 + C2(0) + C3
1 = C3
Therefore, the particular solution to the given differential equation is:
y = (1/4)x^4 + (C1/2)x^2 + C2x + 1
This solution is continuous for negative infinity to positive infinity and has a horizontal tangent at (0,1).
To find a solution to the given differential equation, we need to integrate the equation twice with respect to x.
Step 1: First Integration
Integrate both sides of the equation with respect to x once:
d^2y/dx^2 = 6x
Integrating once, we obtain:
dy/dx = ∫(6x) dx
dy/dx = 3x^2 + C1, where C1 is an arbitrary constant.
Step 2: Second Integration
Integrate both sides of the equation obtained in Step 1 with respect to x again:
∫(dy/dx) dx = ∫(3x^2 + C1) dx
Integrating, we obtain:
y = ∫(3x^2 + C1) dx
y = x^3 + C1x + C2, where C1 and C2 are arbitrary constants.
Step 3: Applying Initial Conditions
Now we need to use the given information to determine the values of the constants C1 and C2. We know that the graph passes through the point (0,1) and has a horizontal tangent there.
Substituting the point (0,1) into the equation obtained in Step 2, we get:
1 = (0)^3 + C1(0) + C2
1 = C2
Step 4: Obtaining the Final Solution
Substituting the value of C2 into the equation obtained in Step 2, we get the final solution:
y = x^3 + C1x + 1
This solution satisfies the given differential equation, is continuous for negative infinity to positive infinity, and passes through the point (0,1) with a horizontal tangent.
Initial value problem.
f"(x)=6x
f'(x)=3x² + C
f(x)=x³ + Cx + D
(each integration introduces one integration constant: C and D)
Given f(0)=1, and f'(0)=0
f(0) = 1 = 0³+C(0)+D => D=1
f'(0) = 0 = 3(0)² + C => C=0
Therefore
f(x)=x³+1
Since the function is a polynomial, its domain is (-∞,∞).