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

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SUMMARY

The discussion focuses on solving the equation \(6\int_{1}^{x} f(t)\, dt + 5 = 3x f(x) - x^3\) for a differentiable function \(f\) defined on the interval \([1, \infty)\) with the initial condition \(f(1) = 2\). The first step involves finding the value of \(f(2)\) and the Laplace transform \(\mathcal{L}\{f(t)\}\). A correction was made regarding the equation, confirming that the correct form is \(6\int_{1}^{x} f(t)\, dt + 5 = 3x f(x) - x^3\). The derived ordinary differential equation (ODE) is \(\frac{d}{dx} f(x) = \frac{f(x)}{x} + x\).

PREREQUISITES
  • Differential equations, specifically first-order linear ODEs
  • Understanding of Laplace transforms, denoted as \(\mathcal{L}\{f(t)\}\)
  • Fundamentals of integral calculus, particularly definite integrals
  • Initial value problems and their solutions
NEXT STEPS
  • Solve the first-order linear ODE \(\frac{d}{dx} f(x) = \frac{f(x)}{x} + x\)
  • Calculate the value of \(f(2)\) using the derived function from the ODE
  • Explore the properties and applications of Laplace transforms in solving differential equations
  • Review techniques for solving integral equations and their relationship with differential equations
USEFUL FOR

Mathematicians, students studying calculus and differential equations, and anyone interested in solving integral equations and their applications in mathematical analysis.

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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 $\mathcal{L} \{ f(t)\}$.
 
Last edited:
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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

$\chi$ $\sigma$
 
Last edited:

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