How to find the general solution of this differential equation?

4 Answers

1. +3Votes  

   xy”=y’+x2e^x,xy"-y’=x2e^x.

   Homogeneous equation: xy”-y’=0.

   dy’/y’=dx/x,ln|y’|=ln|c1x|.

   y’=c1x, y=c1x+c2.

   Set the equation solution y*=(ax+b)e^x.

   y*’=[ax+(a+b)]e^x,

   y*”=[ax+(2a+b)]e^x.

   Substitute the original equation

   [ax2+(2a+b)x]e^x

   =[ax+(a+b)]e^x+ X2e^x.

   So a=1, b=-1.

   y=c1x+c2+(x-1)e^x.

   Elizabeth Wilson- January 11, 2019 |

2. +3Votes  

   Dividing both sides by x2 gives y ́ ́/xy ́/x2=e^x, or decreasing the order to (y ́/x) ́=e^x, the two sides are obtained once: y ́/x=e^x+C1, that is, y ́=xe^x+C1·x, and then integrate once to obtain a general solution: y=(x-1)e^x+C1x2/2+C2, C1 and C2 are Any constant.

   Kimberly- January 12, 2019 |

3. +8Votes  

   Solution: Set the general solution to y=C(x)e^x, then

              y’= C'(x)e^x+C(x)e^x=[C'(x)+C(x)]e^x

              y"=[C"(x)+C'(x)e^x+[C'(x)+C(x)]e^x=[C"(x)+2C'(x)+C( x)]e^x

         Substituting them into the original differential equation, there are

              x[C" (x)+2C'(x)+C(x)]e^x=[C'(x)+C(x)]e^x+x2e^x=[C'(x)+C(x) +x2]e^x

         get x[C"(x)+2C'(x)+C(x)]=C'(x)+C( x)+x2

         ie xC"(x)+(2x-1)C'(x)+(x-1)C(x)=x2

         It can be seen that C(x) is a function. Let C(x)=ax+b, then C'(x)=a, C"(x)=0, substituting into the above formula ,

              (2x-1)a+(x-1)(ax+b)=ax2+(a+b)x-(a+ b)=x2 –> a=1,b=-a=-1

         Therefore C(x)=x-1

         Then  y=C(x)e^x=(x-1)e^x

   Jerry Adams- January 12, 2019 |

4. +3Votes  

   I have seen this question on the exam paper, it seems Is 2006 You look

   Frank Lewis- January 12, 2019 |