differential equation

Question

help me with what method I should use to find the general solution, please

(3x-y+1)y' = -x+3y+5

Answer 1

(3x-y+1)y' = -x+3y+5

(3x-y+1) dy / dx = -x+3y+5

(x−3y−5) d x + (3x-y+1) dy = 0.

Now notice that if the -5 and +1 were not there we would have a nice homogeneous equation.

In addition notice that the lines

x−3y−5 =0

3x-y+1 = 0

intersect at the point ( -1, -2 )

Then the translation x = u - 1 and y = v -2 will change the equations into equations through the origin and since d x = d u and d y = d v the initial equation can be recast in terms of u and v.

( u - 3v ) d u + ( 3u -v ) d v = 0 which is a super nice homogeneous one, which I hope you can do.

Now if we use the substitution v = w u then d v = u d w + w d u.

Therefore the last equation keeps on giving...

( u - 3wu ) d u + (3u - wu )( u d w + w d u. ) =0

( 1 - 3w ) d u + (3 - w )( u d w + w d u. ) =0

( 1 - 3w ) d u + (3 - w )u d w + (3 - w )w d u. =0

( 1 - 3w ) d u + (3 - w )w d u. =− (3 - w )u d w

[1 - 3w + (3 - w )w ] d u = ( u - 3) w d w

[ 1- w² ] d u = ( u - 3) w d w

d u / ( u -3 ) = [ w / (1 - w² )] d w

∫d u / ( u -3 ) = ∫ [ w / (1 - w² )] d w

ln | u -3 | = ( -1/2)ln | w²-1| +k

( u -3 ) √( w²-1) = c and since v = wu

(u - 3 ) √(v²-u²) = c u

and since x = u - 1 and y = v -2 then general solution

( x -2 ) √ ( y²-x² +4y -2x+3 ) = c( x+1) , c constant