Question

linear equation of order one

Answer 1

a,b,n, are constant with n is not equal with 0,n is not equal to -1
(x+a)y'=bx-ny
find general solution of linear equation.

Answer 2

\[\frac{ dy }{ dx }+\frac{ ny }{ x+a}=\frac{ bx }{ x+a }\]

Answer 3

\[(x+a)y'+ny=bx\]divide both sides by (x+a)\[\frac{dy}{dx}+\frac{n}{(x+a)}y=\frac{bx}{(x+a)}\]

Answer 4

i got \[(x+a)^n +c\] for IF

Answer 5

u have gt the right IF

Answer 6

so \[y(x+a)^n=\int\limits_{}^{}\frac{ bx }{ x+a }(x+a)^n\]

Answer 7

i got stuck here, i dont know now lol

Answer 8

R.H.S. \[b \int\limits x(x+a)^{n-1} dx\]\[b \int\limits ((x+a)-a)(x+a)^{n-1}dx\]\[b \int\limits ((x+a)^n-a(x+a)^{n-1})dx\]\[b \int\limits (x+a)^ndx - b \int\limits a(x+a)^{n-1}dx\]

Answer 9

can u solve it further ?

Answer 10

ill try

Answer 11

ill try\[b\frac{ (x+a) ^{2n}}{ 2 }-ab \frac{ (x+a)^{2(n-1)} }{ 2 }\]

Answer 12

no

Answer 13

\[b \frac{(x+a)^{n+1}}{(n+1)}-ab \frac{(x+a)^n}{n}\]

Answer 14

\[b \int\limits (x+a)^n dx\]assume (x+a)=u then dx=du \[b \int\limits u^ndu=b \frac{u^{n+1}}{(n+1)}=b \frac{(x+a)^{n+1}}{(n+1)}\]

Answer 15

gt it now ?

Answer 16

@Caerus

Answer 17

yip

Answer 18

\[y(x+a)^n=\frac{ b(x+a)^{n+1} }{ (n+a)}-\frac{ ab(x+a)^n }{ n }\]

Answer 19

+c

Answer 20

i think this is enough as the answer..

Answer 21

yeah just correct RHS \[\frac{b(x+a)^{n+1}}{(n+1)}-\frac{ab(x+a)^n}{n}+c\]

Answer 22

yehey, thanks again, lets move on