AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

The solution is given as,
y=A(x)e−∫Pdx
The above equation is linear differential equation.
The integral factor is,
IF=e∫P(x)dx
The solution is,
y(IF)=∫Q(x)×IF‌d‌x
ye∫P(x)dx=∫Q(x)e∫P(x)dxdx
Substitute the values in the above equation from given expression
A(x)=∫Q(x)e∫P(x)dxdx
A′(x)=Q(x)e∫P(x)dx