高数,图中的填空第一题怎么写?(答案在横线上了)
1个回答
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设 g(x) 满足 g'(x) = f(x)
f(x) = ∫(0->2x) f(t/2)dt + e
f(x) = 2∫(0->2x) f(t/2)d(t/2) + e
f(x) = 2[g(2x/2)-g(0)] + e
f(x) = 2[g(x)-g(0)] + e
两边取微分:
f'(x) = 2dg(x)/dx = 2 f(x)
df'(x)/dx = 2 f(x)
df'(x)/f(x) = 2 dx
ln(f(x)) = 2x + C
f(x) = Ae^2x
将 f(x) =Ae^2x 代入原方程
Ae^2x = Ae^2x - A + e
A = e
f(x) = e * e^2x = e^(2x+1)
f(x) = ∫(0->2x) f(t/2)dt + e
f(x) = 2∫(0->2x) f(t/2)d(t/2) + e
f(x) = 2[g(2x/2)-g(0)] + e
f(x) = 2[g(x)-g(0)] + e
两边取微分:
f'(x) = 2dg(x)/dx = 2 f(x)
df'(x)/dx = 2 f(x)
df'(x)/f(x) = 2 dx
ln(f(x)) = 2x + C
f(x) = Ae^2x
将 f(x) =Ae^2x 代入原方程
Ae^2x = Ae^2x - A + e
A = e
f(x) = e * e^2x = e^(2x+1)
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