MHB Finding $f$ When $6\int_{1}^{x} f(t)\, dt+5=3x \, f(x)-x^3$

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The discussion revolves around finding the function $f$ defined on the interval $[1,\infty)$, given the equation $6\int_{1}^{x} f(t)\, dt + 5 = 3x f(x) - x^3$. The initial condition states that $f(1) = 2$. A typo was acknowledged, clarifying the equation's correct form. Deriving the equation leads to a first-order linear ordinary differential equation, $\frac{d}{dx} f(x) = \frac{f(x)}{x} + x$, which can be solved using standard methods. The key tasks include determining the value of $f(2)$ and finding the Laplace transform of $f(t)$.
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Let $f:[1,\infty)\to [2,\infty)$ be a differentiable function such that $f(1)=2$. If

$$ 6\int_{1}^{x} f(t)\, dt+5=3x \, f(x)-x^3$$

for all $x \geq 1$, then:

1) Find the value of $f(2)$.

2) Find $\mathcal{L} \{ f(t)\}$.
 
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sbhatnagar said:
Let $f:[1,\infty)\to [2,\infty)$ be a differentiable function such that $f(1)=2$. If

$$ 6\int_{1}^{x} f(t)\, dt=3x \, f(x)-x^3$$

for all $x \geq 1$, then:

1) Find the value of $f(2)$.

2) Find $\mathcal{L} \{ f(t)\}$.
Something wrong here? If $x=1$, then the equation $\displaystyle 6\int_{1}^{x} f(t)\, dt=3x \, f(x)-x^3$ becomes $0=5.$
 
Sorry, I made a typo. The equation is

$$6\int_{1}^{x}f(t)dt+5=3xf(x)-x^3$$

I am really sorry about this.(Sadface)
 
sbhatnagar said:
... the equation is...

$$6\int_{1}^{x}f(t)dt+5=3xf(x)-x^3$$

...

Deriving both terms You arrive to the linear first term ODE...

$\displaystyle \frac{d}{d x}\ f(x)= \frac{f(x)}{x} + x$ (1)

… with initial condition' $f(1)=0$ and the solving procedure is 'standard'...

Kind regards

$\chi$ $\sigma$
 
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