The special solution of the differential equation y ′ SiNx = ylny satisfying the initial condition (when x = π / 2, y = e) is? It is better to write out the steps to solve the problem

The special solution of the differential equation y ′ SiNx = ylny satisfying the initial condition (when x = π / 2, y = e) is? It is better to write out the steps to solve the problem

(dy/dx)sinx=ylny
dy/ylny=sinxdx
d(lny)lny=sinxdx
D (the square of half LNY) = - D (cosx)
The original function is the square of (LNY) = - 2cosx + C
When x = π / 2, y = e, then C = 1
So the square of the original function:: (LNY) = - 2cosx + 1

Find the differential equation. Y '- 2XY = Xe ^ (- x ^ 2)

Let the special solution equation be
a-2x=0
a=2x
So the special solution is y = CE ^ x ^ 2
Because there is e ^ (- x ^ 2)
set up
y=(AX+B)e^(-x^2)
y'=Ae^(-x^2)+(AX+B)e^(-x^2)*(-2x)
=Ae^(-x^2)-2x(AX+B)e^(-x^2)
y'-2xy
=Ae^(-x^2)-2x(AX+B)e^(-x^2)-2x(AX+B)e^(-x^2)
=(A-2AX^2-2BX-2AX^2-2BX)e^(-x^2)
=(A-4AX^2-4BX)*e^(-x^2)=xe^(-x^2)
Then a = 0 - 4B = 1
B=-1/4
So the general solution is
y=C1e^x^2-1/4*e^(-x^2)+C2

Find the general solution of Y '+ 2XY = Xe ^ - x ^ 2

First find the solution of the corresponding homogeneous linear equation
y'+2xy=0
Y = CE ^ (- x ^ 2)
Let y = C (x) e ^ (- x ^ 2)
Substituting into the original equation
C'(x)e^(-x^2)=xe^(-x^2)
C'(x)=x
C(x)=x^2/2+C
y=(x^2/2+C)e^(-x^2)