【高数微分方程题目】用适当变量将下面方程化为可分离变量方程,求通解:y'=y^2+2(sinx-1)y+(sinx)^2-2sinx-cosx+1
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![【高数微分方程题目】用适当变量将下面方程化为可分离变量方程,求通解:y'=y^2+2(sinx-1)y+(sinx)^2-2sinx-cosx+1](/uploads/image/z/1827063-63-3.jpg?t=%E3%80%90%E9%AB%98%E6%95%B0%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8B%E9%A2%98%E7%9B%AE%E3%80%91%E7%94%A8%E9%80%82%E5%BD%93%E5%8F%98%E9%87%8F%E5%B0%86%E4%B8%8B%E9%9D%A2%E6%96%B9%E7%A8%8B%E5%8C%96%E4%B8%BA%E5%8F%AF%E5%88%86%E7%A6%BB%E5%8F%98%E9%87%8F%E6%96%B9%E7%A8%8B%2C%E6%B1%82%E9%80%9A%E8%A7%A3%EF%BC%9Ay%27%3Dy%5E2%2B2%28sinx-1%29y%2B%28sinx%29%5E2-2sinx-cosx%2B1)
【高数微分方程题目】用适当变量将下面方程化为可分离变量方程,求通解:y'=y^2+2(sinx-1)y+(sinx)^2-2sinx-cosx+1
【高数微分方程题目】
用适当变量将下面方程化为可分离变量方程,求通解:
y'=y^2+2(sinx-1)y+(sinx)^2-2sinx-cosx+1
【高数微分方程题目】用适当变量将下面方程化为可分离变量方程,求通解:y'=y^2+2(sinx-1)y+(sinx)^2-2sinx-cosx+1
y'=y^2+2(sinx-1)y+(sinx)^2-2sinx-cosx+1=y^2+2(sinx-1)y+(sinx-1)^2-cosx=(y+sinx-1)^2-cosx
即y'+cosx=(y+sinx-1)^2
令u=y+sinx-1,则原微分方程化为du/dx=u^2,通解是-1/u=x+C,回代yu=y+sinx-1,得原微分方程的通解是y=1-sinx-1/(x+C)
设t=y+sinx-1,则y=t+1-sinx,y'=t'-cosx, 原式化为: t'-cosx=t^2-cosx ==>t'=t^2 ==>dt/t^2=dx ==>t=1/(C-x) ==>y=1/(C-x)+1-sinx.
y'=y² + 2(sinx-1)y + (sinx)²-2sinx-cosx+1 = (y + sinx - 1)² - cosx 即 (y + sinx - 1)' = (y + sinx - 1)² d(y + sinx - 1)/(y + sinx - 1)² = dx 积分得 -1/(y + sinx - 1) = x + C y = -sinx + 1 - 1/(x+C) 这是通解