Question

◆50 points question◆ fast friend solve this differential equation....

Answer 1

is a linear differential equation of first order.It is a linear differential equation of first order.

(y−x)dydx=a2(y−x)dydx=a2

dydx=a2(y−x)dydx=a2(y−x)

dxdy=(y−x)a2dxdy=(y−x)a2

dxdy+xa2=ya2dxdy+xa2=ya2

It is a first order linear differential equation of typeIt is a first order linear differential equation of type

dxdy+P1x=Q1 WhereP1 and Q1 both are function of y alone or constant.dxdy+P1x=Q1 WhereP1 and Q1 both are function of y alone or constant.

P1=1a2,Q1=ya2P1=1a2,Q1=ya2

IF=e∫P1dyIF=e∫P1dy

=eya2=eya2

yeya2=∫ya2eya2dy+cyeya2=∫ya2eya2dy+c

Let ya2=tLet ya2=t

dy=a2dtdy=a2dt

yeya2=a2∫tetdt+cyeya2=a2∫tetdt+c

yeya2=a2et(t−1)+cyeya2=a2et(t−1)+c

yeya2=a2eya2(ya2−1)+c

Answer 2

hlo mate your answer