Solve the initial-value problem for as a function of x.(x^2+36)dy/dx=1,y(6)=0
To solve this differential equation, we can start by separating the variables:
(x^2 + 36) dy = dx
Next, integrate both sides:
∫ (x^2 + 36) dy = ∫ dx
This gives us:
y(x) = ln|x| + C
where C is a constant of integration that we haven't yet determined.
To find C, we need to use the initial condition y(6) = 0. Plugging this into our equation:
0 = ln|6| + C
C = -ln|6|
So our final solution is:
y(x) = ln|x| - ln|6| = ln|x/6|
AAAaannndd the bot gets it wrong yet again!
sorry - you cannot integrate ∫ (x^2 + 36) dy = ∫ dx
But you can do
dy = dx/(x^2+36)
y = 1/6 arctan(x/6) + C
since y(6) = 1, we have
1/6 * π/4 + C = 0
C = -π/24
and so
y = 1/6 arctan(x/6) - π/24