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求微分方程y"+2y=x的通解
人气:169 ℃ 时间:2020-03-23 17:23:09
解答
>> y=dsolve('D2y+2*y=x','x')y =x/2 + C5*cos(2^(1/2)*x) + C6*sin(2^(1/2)*x)>> y'=diff(y)y' =2^(1/2)*C6*cos(2^(1/2)*x) - 2^(1/2)*C5*sin(2^(1/2)*x) + 1/2>> y''=diff(y')y'' =- 2*C5*cos(2^(1/2)*x) - 2*C6*s...
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