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

[sbhatnagar]

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 $\mathcal{L} \{ f(t)\}$.

 

[Opalg]

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 $0=5.$

 

[sbhatnagar]

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)

 

[chisigma]

... 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

$\chi$ $\sigma$

 

[CellCoree]

1) To find the value of $f(2)$, we can use the given equation to set up a system of equations. We know that $f(1)=2$, so we can substitute $x=1$ into the equation to get:

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

This simplifies to $5=3f(1)-1$, which means $f(1)=2$. So now we know that $f(2)$ must satisfy the equation:

$6\int_{1}^{2} f(t)\, dt+5=3(2)\, f(2)-2^3$

To solve for $f(2)$, we can use integration by parts to simplify the integral on the left side:

$6\int_{1}^{2} f(t)\, dt = 6\left[f(t)\, t\right]_{1}^{2} - 6\int_{1}^{2} t\, f'(t)\, dt$

Since $f$ is differentiable, we can use the Fundamental Theorem of Calculus to get:

$6\int_{1}^{2} f(t)\, dt = 6\left[f(2)\, 2 - f(1)\, 1\right] - 6\int_{1}^{2} t\, f'(t)\, dt$

Substituting this into our original equation, we get:

$6\left[f(2)\, 2 - 2\right] - 6\int_{1}^{2} t\, f'(t)\, dt + 5 = 6f(2) - 13$

Simplifying further, we get:

$6f(2) - 1 = 6f(2) - 13$

This simplifies to $f(2)=12$. Therefore, $f(2)=12$.

2) To find $\mathcal{L} \{ f(t)\}$, we can use the Laplace transform property that states:

$\mathcal{L} \left\{ \int_{0}^{t} f(\tau)\, d\tau \right\} = \frac{1}{s} \mathcal{L} \{ f(t