JunhaoZ
contestada

if [tex]f'(x)=[f(x)]^2[/tex], and [tex]f(0)=1[/tex], then [tex]f(6)=\frac{1}{n}[/tex] for some integer [tex]n[/tex]. What is [tex]n[/tex]?

Respuesta :

LammettHash

The ODE is separable: We can write

[tex]f'(x)=f(x)^2\implies\dfrac{\mathrm df}{\mathrm dx}=f^2\implies\dfrac{\mathrm df}{f^2}=\mathrm dx[/tex]

Integrating both sides gives

[tex]-\dfrac1f=x+C[/tex]

so that

[tex]f(0)=1\implies-1=C[/tex]

and so

[tex]-\dfrac1f=x-1\implies f(x)=\dfrac1{1-x}[/tex]

Then [tex]f(6)=-\frac15[/tex], making [tex]n=-5[/tex].

Otras preguntas