y'+f'(x)y=f(x)f'(x)求微分方程

Question

y'+f'(x)y=f(x)f'(x)求微分方程

Answer 1

y=e^[-∫P(x)dx]{∫Q(x)e^[∫P(x)dx]dx+C}
P(x)=f'(x) Q(x)=f(x)f'(x)
∫P(x)dx=∫f'(x)dx=f(x)
∫Q(x)e^[∫P(x)dx]dx=∫f(x)f'(x)e^f(x)dx=∫f(x)d[e^f(x)]=f(x)e^f(x)-∫e^f(x)d(f(x))=f(x)e^f(x)-e^f(x)
所以
y=e^[-f(x)][f(x)e^f(x)-e^f(x)+C]