Differential and integral equation.

[Guest2]

Hello! How do I solve the following integral and differential equations

1. $x^2f''(x)-2xf'(x) = x$

2. $\int_0^{x} (1+f(x))\;{dx} = x$

I'm supposed to http://mathhelpboards.com/linear-abstract-algebra-14/linear-subspaces-17957.html all the polynomials that satisfy this, but I can't. (Shake)

 

[topsquark]

Hello! How do I solve the following integral and differential equations

1. $x^2f''(x)-2xf'(x) = x$

2. $\int_0^{x} (1+f(x))\;{dx} = x$

I'm supposed to http://mathhelpboards.com/linear-abstract-algebra-14/linear-subspaces-17957.html all the functions that satisfy this, but I can't. (Shake)

1. Reduce the order: Let g(x) = f'(x), so your equation becomes \(\displaystyle x^2 g'(x) - 2x g(x) = x\). This is first order linear and you can solve it by method of choice. This equation is exact so no need for an integrating factor.

2. \(\displaystyle \int_0^x (1 + f(x))~dx = \int_0^x (1)~dx + \int_0^x f(x)~dx = x\), so \(\displaystyle \int f(x)~dx = 0\) and go from there.

-Dan

 

[Guest2]

1. Reduce the order: Let g(x) = f'(x), so your equation becomes \(\displaystyle x^2 g'(x) - 2x g(x) = x\). This is first order linear and you can solve it by method of choice. This equation is exact so no need for an integrating factor.

2. \(\displaystyle \int_0^x (1 + f(x))~dx = \int_0^x (1)~dx + \int_0^x f(x)~dx = x\), so \(\displaystyle \int f(x)~dx = 0\) and go from there.

-Dan

1. $f(x) = \left\{a_1-\frac{1}{2}x+a_2 x^3: a_1, a_2 \in \mathbb{R} \right\}$

2. $\frac{1}{2} f^2(x) = 0 \implies f(x) = 0.$

Many thanks.

 

[topsquark]

2. $\frac{1}{2} f^2(x) = 0 \implies f(x) = 0.$

\(\displaystyle \int_0^x f(x)~dx \neq \frac{1}{2} f^2(x)\) in general.

If the anti-derivative of f(x) exists (and it will if f(x) is a polynomial as you have stated) then define \(\displaystyle F(x) = \int f(x)~dx\)

Then \(\displaystyle \int_0^x f(x)~dx = F(x) - F(0)\). So now you need to solve F(x) - F(0) = 0 where F(x) is a polynomial.

Doing this the "long" way we can do the following:
F(x) - F(0) = 0

F(x) = 0 + F(0) = F(0) = constant = k

This means you have F(x) = k and \(\displaystyle F'(x) = f(x) = \frac{d}{dx}(k) = 0\).

-Dan