I'm having a hard time answering this question:
Find all the differentiable functions $f$ such that $$\forall x,y\in\mathbb R : f\left(x^4+y\right) = x^3f(x) + f(y) \text.$$
I would love to get a hint about how to start since this is my first time answering this type of questions.
Thanks a lot.
Differentiate $f(x^4+y)$ by $x$, to get
$$\partial_x f(x^4+y)=4x^3f'(x^4+y)\overset!=\partial_x(x^3f(x)+f(y))=3x^2f(x)+x^3f'(x)$$
Now note that this does not depend on $y$. If $x\neq0$ and you chose $y=z-x^4$ for some $z$ you find $$f'(z)=\frac{3x^2f(x)+x^3f'(x)}{4x^3}$$ so $f'(z)$ is some expression that doesn't even depend on $z$. This means that it has to be constant, and it follows $f(x)=ax+b$ for some constants $a,b$. If you plug it into the initial condition you will find out what these constants are.
You can show that a function $ f : \mathbb R \to \mathbb R $ satisfies $$ f \left( x ^ 4 + y \right) = x ^ 3 f ( x ) + f ( y ) \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $ iff there is a constant $ a \in \mathbb R $ such that $ f ( x ) = a x $ for all $ x \in \mathbb R $, even without the assumption of differentiability. It's easy to check that functions of this form are solutions. We try to prove the converse.
Put $ y = 0 $ in \eqref{0} to get $ f \left( x ^ 4 \right) = x ^ 3 f ( x ) + f ( 0 ) $, which in particular for $ x = 1 $ shows that $ f ( 0 ) = 0 $ and hence $$ f \left( x ^ 4 \right) = x ^ 3 f ( x ) \text . \tag 1 \label 1 $$ This lets you rewrite \eqref{0} as $$ f \left( x ^ 4 + y \right) = f \left( x ^ 4 \right) + f ( y ) \text . $$ This means that for all $ x \in \mathbb R ^ { 0 + } $ and all $ y \in \mathbb R $ we have $ f ( x + y ) = f ( x ) + f ( y ) $. Noting that for any $ x \in \mathbb R $ we have $ | x | \ge 0 $ and $ x + | x | \ge 0 $, we can generalize this and get $$ f ( x + y ) = f \big( ( | x | + x ) + ( y - | x | ) \big) = f ( | x | + x ) + f ( y - | x | ) \\ = f ( | x | ) + f ( x ) + f ( y - | x | ) = f ( x ) + f \big( | x | + ( y - | x | ) \big) \text , $$ and thus $$ f ( x + y ) = f ( x ) + f ( y ) \tag 2 \label 2 $$ for all $ x , y \in \mathbb R $. Letting $ y = - x $ in \eqref{2} we have $$ f ( - x ) = - f ( x ) \tag 3 \label 3 $$ and using induction and \eqref{2} we get $$ f ( n x ) = n f ( x ) \tag 4 \label 4 $$ for every positive integer $ n $. Defining $ a = f ( 1 ) $, we can use \eqref{1} and \eqref{2} to get $$ f \left( ( x + 1 ) ^ 4 \right) = ( x + 1 ) ^ 3 f ( x + 1 ) = ( x + 1 ) ^ 3 \big( f ( x ) + a \big) \text . \tag 5 \label 5 $$ Again, by \eqref{1}, \eqref{2} and \eqref{4} we have $$ f \left( ( x + 1 ) ^ 4 \right) = f \left( x ^ 4 + 4 x ^ 3 + 6 x ^ 2 + 4 x + 1 \right) \\ = f \left( x ^ 4 \right) + f \left( 4 x ^ 3 \right) + f \left( 6 x ^ 2 \right) + f ( 4 x ) + a \\ = x ^ 3 f ( x ) + 4 f \left( x ^ 3 \right) + 6 f \left( x ^ 2 \right) + 4 f ( x ) + a \text . \tag 6 \label 6 $$ Combining \eqref{5} and \eqref{6} we get $$ \left( x ^ 3 + 3 x ^ 2 + 3 x + 1 \right) \big( f ( x ) + a \big) = x ^ 3 f ( x ) + 4 f \left( x ^ 3 \right) + 6 f \left( x ^ 2 \right) + 4 f ( x ) + a \text . \tag 7 \label 7 $$ Similarly, calculating $ f ( x - 1 ) $ in two different ways, and this time also using \eqref{3}, we get $$ \left( x ^ 3 - 3 x ^ 2 + 3 x - 1 \right) \big( f ( x ) - a \big) = x ^ 3 f ( x ) - 4 f \left( x ^ 3 \right) + 6 f \left( x ^ 2 \right) - 4 f ( x ) + a \text , $$ which together with \eqref{7} yields $$ \left( x ^ 3 + 3 x \right) f ( x ) + a \left( 3 x ^ 2 + 1 \right) = x ^ 3 f ( x ) + 6 f \left( x ^ 2 \right) + a \text , $$ or simply $$ f \left( x ^ 2 \right) = \frac 1 2 x f ( x ) + \frac a 2 x ^ 2 \text . $$ Similar to before, we calculate $ f \left( ( x + 1 ) ^ 2 \right) $ in two ways: $$ f \left( ( x + 1 ) ^ 2 \right) = \frac 1 2 ( x + 1 ) \big( f ( x ) + a \big) + \frac a 2 ( x + 1 ) ^ 2 \text ; $$ $$ f \left( ( x + 1 ) ^ 2 \right) = f \left( x ^ 2 \right) + f ( 2 x ) + a = \frac 1 2 x f ( x ) + \frac a 2 x ^ 2 + 2 f ( x ) + a \text ; $$ and these together simplify to $ f ( x ) = a x $.
I haven't done a detailed analysis and proof, but on first look I am finding that a pure linear (not affine) function $f(u) = au$ where $a \in R$ is about the only one that will fit such a strict condition.
If $f(u) = au$, then $f(x^4 + y) = a(x^4 + y) = a x^4 + ay = x^3 * ax + ay = x^3 f(x) + f(y)$
If you can find anything else that works let me know.
Hint.
$$ \lim_{x\to0}\frac{f(y+x^4)-f(y)}{x^4}=\lim_{x\to 0}\frac{f(x)}{x} $$
now as
$$ \lim_{x\to0}\frac{f(y+x^4)-f(y)}{x^4} = f'(y) $$
does not depend on $x$ we have that $\lim_{x\to 0}\frac{f(x)}{x} = C_0 = f'(y)$
The equation is linear, so if $f(x)$ is a solution, so is $\lambda f(x)$. Then by inspection, $f(x)=x$ works.
Let $h:=x^4$. We have
$$f'(y)=\lim_{h\to0^+}\frac{f(y+h)-f(y)}h=\lim_{h\to0^+}\frac{h^{3/4}f(h^{1/4})}h=f'(0)$$ provided $$f(0)=0$$ and by integration $$f(y)=f'(0)y=cy.$$