Umwege

Algebraischer Lösungsansatz für f '(x) = f(x) + x

Transformation g(x) = x + f(x) $\Rightarrow$ g ' (x) = 1 + f '(x) = 1+ x + f(x) = 1 + g(x)

Wir untersuchen also die Differenzialgleichung

g'(x) = 1 + g(x)

Separation: $\frac{g'(x)}{1+g(x)}= 1 \Rightarrow$ ln(1+g(x)) = x + C $\Rightarrow$ e^x+C = 1 + g(x) $\Rightarrow$

g(x) = ae^x -1 = x + f(x) $\Rightarrow$

f(x) = ae^x - x - 1

Probe: f '( x) = ae^x - 1 = f(x) + x