题目

求给定微分方程的特解
求微分方程满足所给初始条件的特解y'+x^2* y=x^2 ,当x=2,y =1
我解得：x=2时，Y=1。代进去后成 LN Y-1=LN0了，这个没解了

提问时间：2021-03-23

答案

dy/dx+(x^2)y=x^2对应齐次方程为:dy/dx+(x^2)y=0dy/y=-(x^2)dxIny=-(x^3)/3+InCIn(y/C)=-(x^3)/3y=Ce^[-(x^3)/3]=C(x)e^[-(x^3)/3]dy/dx=C'(x)e^[-(x^3)/3]-(x^2)C(x)e^[-(x^3)/3]代入原方程C'(x)e^[-(x^3)/3]-(x^2)...