Solve the inital value problem for y as a function of x given: dy/dx=2x-6 with an inital condition of y=4 when x=1.
antiderivative->x^2-6x+c
4=1^2-6(1)+c
c=9
y-x^2-6x+9
see previous solution
Do you mean c=9?
yes, your solution is correct, except write it up this way:
dy/dx = 2x-6
y = x^2 - 6x + c and then proceed as you did before
I didn't realize I put y- instead of y=. So, you mean write it y=x^2-6x+9?
To solve the initial value problem, we need to find the function y(x) that satisfies the differential equation dy/dx = 2x - 6 and passes through the point (1, 4).
First, we need to find the antiderivative of the right-hand side of the equation. The antiderivative of 2x - 6 with respect to x is x^2 - 6x + C, where C is the constant of integration.
Next, we substitute the initial condition y = 4 when x = 1 into the antiderivative.
4 = 1^2 - 6(1) + C
Simplifying, we have:
4 = 1 - 6 + C
4 = -5 + C
C = 9
Now, we have the constant of integration C as 9. Thus, the solution to the initial value problem is:
y = x^2 - 6x + 9