Solved Problems on de With Homogeneous Coefficient

Solved Problems on Differential Equations with Homogeneous Coefficient prof-desk.com Matthew Suan

The methods and techniques used in this document to solve differential equations with homogeneous coefficients are already discussed in the article ”Differential Equations with Homogeneous Coefficients” by the same author. So before attempting to go through the following examples, make sure you read the above mentioned article or else have basic knowledge in solving differential equations with homogeneous coefficients. However, if you feel like you can go through directly, you may proceed. :-) Enjoy! Example 1 Solve the following differential equation.

x2 + 2xy − 4y 2 dx − x2 − 8xy − 4y 2 dy = 0.

(1)

We know that the coefficients of the above differential equation are both homogeneous of order 2. Thus, we can make the necessary substitution y = vx or x = vy. In doing so, we let y = vx such that dy = vdx + xdv. Inserting this into (1), we get

x2 + 2x(vx) − 4(vx)2 dx − x2 − 8x(vx) − 4(vx)2 (vdx + xdv) = 0. (2) Simplifying (2), we have

x2 1 + 2v − 4v 2 dx − x2 1 − 8v − 4v 2 (vdx + xdv) = 0. 1

(3)



1 + 2v − 4v 2 dx − 1 − 8v − 4v 2 (vdx + xdv) = 0.

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1 + v + 4v 2 + 4v 3 dx − x 1 − 8v − 4v 2 dv = 0.

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We can now readily separate variables to get dx (1 − 8v − 4v 2 ) − dv = 0. x (1 + v + 4v 2 + 4v 3 )

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We can now integrate with the help of standard tables of integration (1 − 5v + v 2 ) dv = 0. (1 + 4v + 3v 2 − v 3 )

(7)

ln x = log (v + 1) − log (8v 2 + 2) = c.

(8)

Z

dx − x

Z

which yields

Substituting back v = xy , we finally get the solution which is x2 + 4y 2 = c(y + x).

(9)

Remark: Note that we use the option y = vx in making the necessary substitution. However, we can also use x = vy and and should get the same answer to the DE in (1) which is shown in equation (9). Example 2 Solve the following differential equation.   y x cos2 − y dx + xdy = 0 x

(10)

Upon inspection, we can see that the differential equation in (10) is homogeneous of degree 1. Thus, we can make substitution to transform the differential equation into a variable separable differential equation. The choice of variable for the necessary substitution is important here. I would suggest that we use y = vx such that when we its derivative dy = vdx + xdv is inserted into the second term of equation (10), we only have 2

one multiplier which is x thereby making the algebraic simplification much easier. Thus, letting y = vx, equation (10) is transformed into 

x cos2

 vx − vx dx + x(vdx + xdv) = 0 x

(11)

which is just
x cos2 v − vx dx + x(vdx + xdv) = 0.

(12)

We then cancel the common term which is x to get
cos2 (v) − v dx + vdx + xdv = 0

(13)

cos2 vdx + xdv = 0

(14)

Thus, we now have transformed (10) into a variable saparable differential equation in (14). Proceeding accordingly, we get dv dx + =0 x cos2 v

(15)

which is just integrated in the form of Z

dx + x

Z

Z

2

sec vdv =

0

(16)

and yields ln |x| + tan v = c

(17)

Putting back the assumption y = vx, we obtain ln |x| + tan

y =c x

(18)

which is now the general solution to the homogeneous differential equation in (10). Again, we can still chose to substitute x = yv and should get us the same answer. 3

Example 3 Solve the general solution of the following differential equation. (x − y ln y + y ln x)dx + x(ln y − ln x)dy = 0.

(19)

If we look at the equation directly, you might say that the coefficients are not homogeneous. But knowing that ln a − ln b = ln ab and ln a + ln b = ln ab, we can rewrite (19) into    y x dx + x ln dy = 0. x − y ln y x

(20)

which is now a differential equation whose coefficients are homogeneous of the first order. Thus, we let y = vx and dy = vdx + xdv to transform equation (19) into 

x − vx ln

 vx  x dx + x ln (vdx + xdv) = 0. vx x

(21)

We then cancel the common term x to get 

1 1 − v ln v

 dx + (ln v) (vdx + xdv) = 0.

(22)

which is rewritten into (1 + v ln v) dx + (ln v) (vdx + xdv) = 0.

(23)

We then add the similar terms to obtain (1 + 2v ln v) dx + x ln vdv = 0.

(24)

Now, the DE above is separable that it becomes dx ln v + dv = 0. x 1 + 2v ln v

(25)

Thus, we have successfully separated the variables and the solution to the differential equation in (19) is found by integrating, Z

dx + x

Z

ln v dv = 1 + 2v ln v 4

Z 0.

(26)

which is just (x − y) ln |x| + y ln y = cx + y.

(27)

Example 4 Solve the general solution of the following differential equation. x2 y 0 = 4x2 + 7xy + 2y 2 .

(28)

As we usually did, we let y = vx such that x2 dy = (4x2 + 7xy + 2y 2 )dx

(29)

x2 (vdx + xdv) = (4x2 + 7x(vx) + 2(xv)2 )dx

(30)

becomes

from which we cancel the common terms to get vdx + xdv = (4 + 7v + 2v 2 )dx.

(31)

xdv = (4 + 6v + 2v 2 )dx.

(32)

Rearranging, we have

Separating variables, we then have dx dv = . 4 + 7v + 2v 2 x

(33)

From which integrate Z

dv = 4 + 7v + 2v 2

Z

dx . x

(34)

which will then yield a result of ln | − v − 1| − ln |v + 2| + c = 2 ln x

5

(35)

Now, we will put back y = vx into the equation above and do some algebraic manipulation to obtain x2 (y + 2x) = c(y + x). This is now the general solution to our original problem.

6

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