let f be a function such that y'[x^2+1*f(x)]=2xf(x) +3x^4 +3x^2. What's f(x)?
To determine the function f(x) from the given differential equation, we need to solve for it step by step. Here's how to find f(x):
Step 1: Rewrite the equation to isolate the derivative term y'[x^2 + 1 * f(x)]:
y'[x^2 + f(x)] = 2xf(x) + 3x^4 + 3x^2
Step 2: Differentiate both sides with respect to x:
(derivative of y'[x^2 + f(x)]) = (derivative of 2xf(x) + 3x^4 + 3x^2)
Step 3: Use the chain rule on the left side. Let u = x^2 + f(x), so du/dx = 2x + f'(x):
y''[x^2 + f(x)] * (2x + f'(x)) = 2xf(x) + 3x^4 + 3x^2
Step 4: Expand the left side and simplify the equation:
2x * y''[x^2 + f(x)] + f'(x) * y''[x^2 + f(x)] = 2xf(x) + 3x^4 + 3x^2
Step 5: It is often convenient to introduce a new variable for the derivative y''[x^2 + f(x)], so let y''[x^2 + f(x)] = g(x). Now we have:
2x * g(x) + f'(x) * g(x) = 2xf(x) + 3x^4 + 3x^2
Step 6: Rearrange the equation to get f'(x) in terms of g(x) and f(x):
f'(x) = (2xf(x) + 3x^4 + 3x^2) / g(x) - 2x
Step 7: Integrate both sides with respect to x to find f(x):
∫ f'(x) dx = ∫ [(2xf(x) + 3x^4 + 3x^2) / g(x) - 2x] dx
f(x) = ∫ [(2xf(x) + 3x^4 + 3x^2) / g(x) - 2x] dx + C
Where C is the constant of integration.