Question

求微积分大神Let f and g be functions that are differentiable for a

Answer 1

(a)f'(x)+g'(x)=f(x)-g(x)+g(x)-f(x)=0

so [f(x)+g(x)]'=0

f(x)+g(x)=C

f(0)+g(0)=C=5+1=6

sof(x)+g(x)=6

(b)plug in g(x)=6-f(x) to the first equation

f'(x)=f(x)-(6-f(x))

f'(x)=2f(x)-6

那就分离变量

df/dx=2f-6

df/(f-3)=2dx

两边积分

ln(f-3)=2x+C'

f-3=e^(2x+C')

f=3+Ce^(2x)

代入x=0,f=5

C=2

f=3+2e^(2x)

g=6-f=3-2e^(2x)

f'(x)-2f(x)=-6

multiply both sides by e^(-2x) (积分因子)

e^(-2x)f'(x)-2e^(-2x)f(x)=-6

左边等价于

e^(-2x)f'(x)+(e^(-2x))'f(x)=(e^(-2x)f(x))' (导数的积法则倒过来)

(e^(-2x)f(x))'=-6e^(-2x)

so

e^(-2x)f(x)=∫-6e^(-2x)dx

=3e^(-2x)+C

f(x)=3+Ce^(2x)

plug in x=0,f(0)=5

3+C=5,C=2

so

f(x)=3+2e^(2x)

g(x)=3-2e^(2x)