matlab解二阶微分方程
怎么用matlab来解呢?x的定义域是(0,l),i是道路横坡坡度,w是降雨强度,k是路面横向渗透系数。...
怎么用matlab来解呢?
x的定义域是(0,l),i是道路横坡坡度,w是降雨强度,k是路面横向渗透系数。 展开
x的定义域是(0,l),i是道路横坡坡度,w是降雨强度,k是路面横向渗透系数。 展开
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你的方程即为:
y'*y' + y''*y' +w/k =0
s = dsolve('Dy*D2y + Dy^2 + w/k ', 'y(l) = a', 'Dy(0) = i1','x'); % i1即i
用2011b可得如下结果:
a + (exp(log(k*i1^2 + w) - 2*l) - w)^(1/2)/k^(1/2) - (exp(log(k*i1^2 + w) - 2*x) - w)^(1/2)/k^(1/2) - (w^(1/2)*atan(((k*i1^2 + w)/exp(2*l) - w)^(1/2)/w^(1/2)))/k^(1/2) + (w^(1/2)*atan(((k*i1^2 + w)/exp(2*x) - w)^(1/2)/w^(1/2)))/k^(1/2)
a - (exp(log(k*i1^2 + w) - 2*l) - w)^(1/2)/k^(1/2) + (exp(log(k*i1^2 + w) - 2*x) - w)^(1/2)/k^(1/2) + (w^(1/2)*atan(((k*i1^2 + w)/exp(2*l) - w)^(1/2)/w^(1/2)))/k^(1/2) - (w^(1/2)*atan(((k*i1^2 + w)/exp(2*x) - w)^(1/2)/w^(1/2)))/k^(1/2)
y'*y' + y''*y' +w/k =0
s = dsolve('Dy*D2y + Dy^2 + w/k ', 'y(l) = a', 'Dy(0) = i1','x'); % i1即i
用2011b可得如下结果:
a + (exp(log(k*i1^2 + w) - 2*l) - w)^(1/2)/k^(1/2) - (exp(log(k*i1^2 + w) - 2*x) - w)^(1/2)/k^(1/2) - (w^(1/2)*atan(((k*i1^2 + w)/exp(2*l) - w)^(1/2)/w^(1/2)))/k^(1/2) + (w^(1/2)*atan(((k*i1^2 + w)/exp(2*x) - w)^(1/2)/w^(1/2)))/k^(1/2)
a - (exp(log(k*i1^2 + w) - 2*l) - w)^(1/2)/k^(1/2) + (exp(log(k*i1^2 + w) - 2*x) - w)^(1/2)/k^(1/2) + (w^(1/2)*atan(((k*i1^2 + w)/exp(2*l) - w)^(1/2)/w^(1/2)))/k^(1/2) - (w^(1/2)*atan(((k*i1^2 + w)/exp(2*x) - w)^(1/2)/w^(1/2)))/k^(1/2)
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