Question

Solve y'=(2x+y+1)/(x+2y-4) implicitly.

Answer 1

@myininaya

Answer 2

So dy/dx=(2x+y+1)/(x+2y-4), and then?

Answer 3

Please help me.

Answer 4

is it homogeneous ?

Answer 5

if not, can you make it homegeneous forcibly ?

Answer 6

because you can always solve a homogeneous equation

Answer 7

y=ux
u=y/x
y'=u'x+u
y'=(2+y/x+1/x)/(1+2(y/x)-4/x)
=(1/x)(2x+y+1)/(1/x)(x+2y-4)
=(2x+y+1)/(x+2y-4)

Answer 8

making a linear equation homogeneous is easy,
all you need to is to shift it up/down :

Answer 9

that substitution works only with homogeneous equations @Idealist10
so think of a way to change the given equation into homogeneous form

Answer 10

those constants +1 and -4 are making this equation difficult, right ?

Answer 11

Yes.

Answer 12

lets get rid off them

y'=(2x+y+1)/(x+2y-4)

say \(\large u = x+h\)
\(\large v = y+k\)

\[\large \dfrac{dy}{dx} = \dfrac{dv}{du} \]

the given equation changes to :

\[\large \dfrac{dv}{du} = \dfrac{2(u-h) + v-k+1}{u-h+2(v-k)-4} \]

Answer 13

choose the values of "h" and "k" such that they eat off the constant terms

Answer 14

a system of equations?

Answer 15

simplifying a bit you get


\[\large \dfrac{dv}{du} = \dfrac{2u + v + (\color{red}{-2h-k+1})}{u+2v + (\color{red}{-h-2k-4})} \]

Answer 16

exactly setting those constant terms equal to 0 gives a system of two linear equations which we can solve for good h,k values

Answer 17

it seems \(\large h=2, k=-3 \) are the magic numbers

that choice of h,k kills the constant terms and you're left with your favorite homogeneous equation to solve :

\[\large \dfrac{dv}{du} = \dfrac{2u + v}{u+2v } \]

Answer 18

solve it,
we can change back to xy coordinates after solving this

Answer 19

@myininaya