求方程的通解求这个方程的通解,跪谢
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求方程的通解
求这个方程的通解,跪谢
![](http://img.wesiedu.com/upload/6/1a/61ab125b73b3d52277ce943365813bd8.jpg)
求这个方程的通解,跪谢
![](http://img.wesiedu.com/upload/6/1a/61ab125b73b3d52277ce943365813bd8.jpg)
![求方程的通解求这个方程的通解,跪谢](/uploads/image/z/19032372-36-2.jpg?t=%E6%B1%82%E6%96%B9%E7%A8%8B%E7%9A%84%E9%80%9A%E8%A7%A3%E6%B1%82%E8%BF%99%E4%B8%AA%E6%96%B9%E7%A8%8B%E7%9A%84%E9%80%9A%E8%A7%A3%2C%E8%B7%AA%E8%B0%A2)
y"/y' = 1/y' *d(y')/dx = 1/x + 2x
d(y')/y' = dx/x + 2x*dx
两边同时积分,可以得到:
ln(y') = lnx + x^2 + c1
则 y'=dy/dx = C'* x * e^(x^2),C'= e^(c1)
dy = C'* x * e^(x^2) * dx = (C'/2) * e^(x^2) * (2x*dx) = (C'/2) * e^(x^2) * d(x^2)
所以,
y = (C'/2) * e^(x^2) + C"
d(y')/y' = dx/x + 2x*dx
两边同时积分,可以得到:
ln(y') = lnx + x^2 + c1
则 y'=dy/dx = C'* x * e^(x^2),C'= e^(c1)
dy = C'* x * e^(x^2) * dx = (C'/2) * e^(x^2) * (2x*dx) = (C'/2) * e^(x^2) * d(x^2)
所以,
y = (C'/2) * e^(x^2) + C"