Consider the following Differential equation: (2x + y + 4)dx= (x - 2y - 2)dy 1.

Question

Consider the following Differential equation: (2x + y + 4)dx= (x - 2y - 2)dy 1. Write the equation in the form (a_1 x + b_1y + c_1) dx + (a_2x + b_2y + c_2) dy = 0 2. Solve the system of equation a_1 h + b_1k + c_1 = 0 a_2h + b_2k + c_2 = 0 To find the values of the constants h and k 3. Use the substitution x = u + h and y = v + k to transfer the equation into the form dv/du = -a_1u+b_1v/a_2u+b_2v = -a_1 + b_1(v/u)/a_2 + b_2 (v/u) Which is homogeneous. 4. Solve the homogeneous equation in (3) with respect u and v. 5. Re-substitute u = x - h and v = y - k to write your solution in terms of x and y. 6. Discussion and conclusion. Best Wishes

Explanation / Answer

1. The equation can be written as

(2x+y+4) dx + (-x+2y+2) dy = 0

Here a1 = 2 b1 = 1 and c1 = 4 ; a2 = -1 b2 = 2 and c2 = 2

2. The system of equations is

2h + k + 4 = 0 (1)

-h + 2k + 2 = 0 (2)

Multiplying (1) by 2 and subtracting (2) we have

(4h+h) + (2k-2k) + (8-2) = 0

=> 5h + 6 = 0

=> h = -6/5

Substituting in (1),

2*(-6/5) + k + 4 = 0

-12/5 + k + 4 = 0

=> k = -4 + 12/5

=> k = -8/5

3. x = u + h = u - 6/5

y = v + k = v - 8/5

dx = du and dy = dv

Thus the given equation becomes

=> (2(u-6/5)+(v-8/5)+4) du + (-(u-6/5)+2(v-8/5)+2) dv= 0

=> (2u - 12/5 + v - 8/5 + 4) du + (-u + 6/5 + 2v - 16/5 +2) dv = 0

=> (2u + v) du + (-u +2v) dv = 0

=> dv/du = - (2u+v) / (-u+2v) = - (2 + v/u) / (-1 + 2v/u)

4. Let z = v/u

=> v = uz

=> dv/du = u * dz/du + z * du/du

=> dv/du = u * dz/du + z

Substituting in the homegenous equation

=> u (dz/du) + z = - (2+z) / (-1+2z)

=> dz/du = [- (2+z) / (-1+2z) - z] / u

=> dz/du = - (2+z+z-2z2) / [(-1+2z) u]

=> du/u = -dz (2z-1) / ( 2z2-2z-2)

=> (2/u) du = -[(2z-1)/(z2-z-1)] dz

Integrating both sides,

=> 2 log u = -log (z2-z-1)

=> log u2 = -log (z2-z-1)

=> u2 = 1/(z2-z-1)

=> u2 = 1/ [(v/u)2 - (v/u) - 1]

=> u2 ([ v2 - vu - u2] / u2) = 1

=> v2 -vu - u2 = 1

5. Substituting u = x - h and v = y - k

=> (y+8/5)2 - (y+8/5) (x+6/5) - (x+6/5)2 -1 = 0

=> y2 + 16y/5 + 64/25 - xy - 48/25 -8x/5 -6y/5 - x2 - 36/35 - 12x/5 - 1 = 0

=> y2 - x2 -4x +2y -xy -11/7 = 0.

6. Thus in order to solve a non homogenous equation of the form (a1x + b1y + c1) dx = (a2x + b2y + c2) dy,

we try to eliminate the constants c1 and c2 with suitable substitutions x = u + h and y = v + k.

We then solve the resulting homogenous equation with the usual tactics before we reverse substitute the parameters u and v.

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