if f be a function such that d/dx[x^2+1)f(x)]=2xf(x)+3x^4+3x^2
what is f(x)?
To find the function f(x), we need to solve the given differential equation:
(d/dx)[x^2 + 1) * f(x)] = 2xf(x) + 3x^4 + 3x^2
Let's simplify the left-hand side of the equation using the product rule of differentiation:
d/dx[x^2 + 1) * f(x)] = (2x * f(x) + (x^2 + 1) * f'(x)
Now substituting this back into the original equation, we have:
2x * f(x) + (x^2 + 1) * f'(x) = 2xf(x) + 3x^4 + 3x^2
The "2xf(x)" term on both sides of the equation cancels out:
(x^2 + 1) * f'(x) = 3x^4 + 3x^2
Now, let's isolate the derivative term by dividing both sides by (x^2 + 1):
f'(x) = (3x^4 + 3x^2) / (x^2 + 1)
Next, we can integrate both sides of the equation to find f(x). The integral of the right-hand side will require some algebraic manipulation:
∫f'(x) dx = ∫(3x^4 + 3x^2) / (x^2 + 1) dx
The right-hand side can be rewritten as:
= ∫[3x^2 - 3 + (3 / (x^2 + 1))]dx
The integral of 3x^2 - 3 can be easily computed as:
= x^3 - 3x
Now, let's focus on the integral (3 / (x^2 + 1)):
To solve this integral, we can perform a substitution u = x^2 + 1:
du = 2x dx
Rearranging, we have:
x dx = (1/2) du
Substituting back into the integral, we get:
= ∫(3 / (x^2 + 1)) dx = (3/2) ∫(1 / u) du = (3/2) ln|u| = (3/2) ln|x^2 + 1| + C
Finally, combining the two results obtained from integrating, we have:
f(x) = x^3 - 3x + (3/2) ln|x^2 + 1| + C
where C is the constant of integration. Therefore, the function f(x) satisfying the given differential equation is:
f(x) = x^3 - 3x + (3/2) ln|x^2 + 1| + C