the velocity v of a particle is given by v=1-4x where x is p

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the velocity v of a particle is given by v=1-4x where x is position.if the initial velocity of the particle is 2m/s,find an equation for x in terms of t.
没有说是匀加速运动

这是道微分方程题目,应该这样做：
设x=f(t)
f'(t)=v=1-x
dx/dt=1-x
也就是说,求初始条件f'(0)=2m/s的微分方程dx/dt=1-x的解.
可用分离变量的微分方程的解法：
(1-x)dx=dt
积分得：
x-x^2/2=t+C1
x^2-2x+t+C1=0
(x-1)^2+t+C2=0 [C2=C1-1]
(x-1)^2=C-t [C=-C2]
x-1=±√(C-t)
x=1±√(C-t)
∴f'(t)=±(C-t)^(-1/2)(-1)
=±1/√(C-t)
∴f'(0)=1/√C=2
C=1/4
∴x=1+√[(1/4)-t]
没想到都扔了20多年了还能解简单的微分方程,也不知对否.

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