1、求微分方程的解析解、并使用MATLAB画出它们的图形 （1）y(4)=y,y(0)=y’(0)=2，y”(0)=y””(0)=1

这是一个四阶常微分方程，可以写成特征方程：

r^4 - 1 = 0

解得：

r = ±1, ±i

因此通解为：

y(t) = c1*e^t + c2*e^(-t) + c3*cos(t) + c4*sin(t)

带入初始条件：

y(0) = c1 + c2 + c3 = 0

y'(0) = c1 - c2 + c4 = 2

y''(0) = c3 + c4 = 1

y'''(0) = -c3 + c4 = 0

解得：

c1 = c2 = c3 = 0, c4 = 1

因此特解为：

y(t) = e^t - e^(-t) + sin(t)

MATLAB代码如下：

syms y(t)
ode = diff(y, t, 4) == y;
cond1 = y(0) == 2;
cond2 = diff(y)(0) == 2;
cond3 = diff(y, t, 2)(0) == 1;
cond4 = diff(y, t, 3)(0) == 1;
conds = [cond1 cond2 cond3 cond4];
ySol(t) = dsolve(ode, conds);
fplot(ySol,[-10,10])