Mathematics Solution of Homogeneous differential equations For CBSE-NCERT
Click for Only Video

### Topic covered

color{red} ♦  Solution of Homogeneous differential equations

### Homogeneous differential equations

Consider the following functions in x and y

color{red} {F_1 (x, y) = y^2 + 2xy, F_2 (x, y) = 2x – 3y},

color{red} {F_3 (x,y) = cos (y/x) , F_4 (x, y) = sin x + cos y}

If we replace x and y by λx and λy respectively in the above functions, for any nonzero constant λ, we get

F_1 (λx, λy) = λ^2 (y^2 + 2xy) = λ^2 F_1 (x, y)

F_2 (λx, λy) = λ (2x – 3y) = λ F_2 (x, y)

F_3 (λx, λy) = cos ((lamday)/(lamdax)) = cos (y/x) = lamda^0 F_3 (x,y)

F_4 ( λx, λy) = sin λx + cos λy ≠ λ^n F_4 (x, y), for any n ∈ N

\color{green} ✍️ Here, we observe that the functions F_1, F_2, F_3 can be written in the form F(λx, λy) = λ^n F (x, y) but F_4 can not be written in this form. This leads to the following definition:

color{blue}{"A function F(x, y) is said to be homogeneous function of degree n if" F(λx, λy) = λ^n F(x, y) "for any nonzero constant λ."}

=> We note that in the above examples, F_1, F_2, F_3 are homogeneous functions of degree 2, 1, 0 respectively but F_4 is not a homogeneous function.

We also observe that

F_1(x,y) = x^2 (y^2/x^2 + (2y)/x) = x^2 h_1 (y/x)

or F_1(x,y) = y^2 ( 1 + (2x)/y) = y^2 h_2 (x/y)

F_2 (x,y) = x^1 (2 - (3y)/x ) = x^1h_3 (x/y)

or F_2 (x,y) = y^1(2 x/y -3) = y^1h_4(x/y)

F_3 (x,y) =x^0 cos (y/x) =x^0 h_5 (y/x)

F_4 (x,y) != x^n h_6 (y/x), for any n in N

or  F_4 (x,y) != y^n h_1 (x/y), for any n in N

=> Therefore, a function F (x, y) is a homogeneous function of degree n if

color{blue}{F(x,y) = x^n g(y/x) \ \ "or" \ \ y^nh(x/y)}

\color{green} ✍️ (dy)/(dx) = F (x, y) is said to be homogenous if F(x, y) is a homogenous function of degree zero.

color{red}{"Simplification Techniques"}

=> To solve a homogeneous differential equation of the type

color{red} {(dy)/(dx) =F(x,y) = g (y/x)} ........................(1)

=> We make the substitution color{red} {y = v . x} ... (2)

=> Differentiating equation (2) with respect to x, we get

color{red} {(dy)/(dx) = v + x (dv)/(dx)} ...............................(3)

=> Substituting the value of (dy)/(dx) from equation (3) in equation (1), we get

v + x (dv)/(dx) = g(v)

or color{red} {x(dv)/(dx) = g(v) - v} ..............................(4)

=> Separating the variables in equation (4), we get

color{red} {(dv)/(g(v) -v) = (dx)/x} .............................(5)

Integrating both sides of equation (5), we get

color{red} {int(dv)/(g(v) -v) = int 1/x dx +C} ....................(6)

Equation (6) gives general solution (primitive) of the differential equation (1) when we replace color{green}v by color{orange}{y/x}

"Key Concept :"
If the homogeneous differential equation is in the form (dx)/(dy) = F(x,y) where, F (x, y) is homogenous function of degree zero, then we make substitution x/y =v i.e., x = vy and we proceed further to find the general solution as discussed above by writing (dx)/(dy) = F(x,y) = h(x/y)
Q 3167245185

Show that the differential equation (x – y) (dy)/(dx) = x +2y is homogeneous and solve it.
Class 12 Chapter 9 Example 15
Solution:

The given differential equation can be expressed as

(dy)/(dx) = (x +2y)/(x -y) ..........................(1)

Let F(x, y) = (x +2y)/(x -y)

Now F(lamda x , lamda y) = (lamda(x +2y))/(lamda (x -y)) = lamda^0 * f(x,y)

Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.

Alternatively,

(dy)/(dx) = ((1 + (2y)/x)/(1 - y/x) ) = g(y/x)

R.H.S. of differential equation (2) is of the form g(y/x) and so it is a homogeneous function of degree zero. Therefore, equation (1) is a homogeneous differential equation.

To solve it we make the substitution

y =vx ........................(3)

Differentiating equation (3) with respect to, x we get

(dy)/(dx) = v + x (dv)/(dx) .....................(4)

Substituting the value of y and (dy)/(dx) in equation (1) we get

v+x (dv)/(dx) = (1 +2v)/(1-v)

or x (dv)/(dx) = (1 +2v)/(1-v) -v

or x(dv)/(Dx) = (v^2 +v +1 )/(1-v)

or (v-1)/(v^2 +v +1) dv = (-dx)/x

Integrating both sides of equation (5), we get

int (v-1)/(v^2 +v +1 ) dv = - int (dx) /x

or 1/2 int (2v +1 -3 )/( v^2 +v +1) dv = - log |x| + C_1

or  1/2 int (2v +1) / (v^2 +v +1) dv - 3/2 int 1(v^2 + v +1 ) dv = - log |x | + C_1

or  1/2 log |v^2 + v +1 | -3/2 int 1/ ((v +1/2 )^2 + (sqrt3/2)^2) dv = - log |x| +C_1

or 1/2 log |v^2 +v +1 | - 3/2 * 2/sqrt3 tan^-1 ((2v + 1) /sqrt3) = - log |x| + C_1

or  1/2 log |v^2 +v + 1 | + 1/2 log x^2 = sqrt3 tan^-1 ((2v +1 )/sqrt3) +C_1 (why ? )

Replacing v by y/x, we get

or 1/2 log | y^2/x^2 + y/x + 1 | + 1/2 log x^2 = sqrt3 tan^-1 ((2y +x)/(sqrt3 x)) +C_1

or  1/2 log | (y^2/x^2 + y/x +1) x^2 | = sqrt3 tan^-1 ((2y +x)/(sqrt3x)) + C_1

or  log | (y^2 +xy +x^2 )| = 2 sqrt3 tan^-1 ( (2y +x ) / ( sqrt3 x )) + 2 C_1

or  log | (x^2 + xy + y^2) | = 2 sqrt3 tan^-1 ( (x +2y)/(sqrt3 x ) ) + C

which is the general solution of the differential equation (1)
Q 3187545487

Show that the differential equation x cos (y/x) (dy)/(dx) = y cos (y/x) + x is homogeneous and solve it.
Class 12 Chapter 9 Example 16
Solution:

The given differential equation can be written as

(dy)/(dx) = (y cos (y/x) + x)/(x cos (y/x)) .................(1)

It is a differential equation of the form (dy)/(dx) = F(x,y)

Here F(x,y) = (y cos (y/x) +x ) /(x cos (y/x) )

Replacing x by λx and y by λy, we get

F(lamdaa, lamday ) = (lamda [ y cos (y/x) + x ])/ (lamda ( x cos \ \y/x )]

Thus, F(x, y) is a homogeneous function of degree zero.

Therefore, the given differential equation is a homogeneous differential equation.

To solve it we make the substitution

y = vx ... (2)

Differentiating equation (2) with respect to x, we get

(dy)/(dx) = v + x (dv)/(dx) .......................(3)

Substituting the value of y and (dy)/(dx)  in equation (1) , we get

v +x (dv)/(dx) = ( v cos v + 1)/(cos v)

or  x (dv)/(dx) = ( v cos v + 1) /(cos v) - v

or  x (dv)/(dx ) = 1/(cos v )

or  cos v dv = (dx )/x

Therefore  int cos v dv = int 1/x dx

or sin v = log | x | + log |C|

or sin v = log |Cx |

Replacing v by y/x , we get

sin (y/x) = log |Cx|

which is the general solution of the differential equation (1).
Q 3187756687

Show that the differential equation 2ye^(x/y) dx + ( y - 2xe^(x/y) )dy = 0 is homogeneous and find its particular solution, given that, x = 0 when y = 1.
Class 12 Chapter 9 Example 17
Solution:

The given differential equation can be written as

(dx)/(dy) = (2x e^(x/y) - y)/(2ye^(x/y)) ............................(1)

Let F(x,y) = ( 2xe^(x/y) -y )/(2ye^(x/y))

Then F(lamda x, lamda y) = ( lamda ( 2xe^(x/y) - y ))/(lamda( 2 y e^(x/y))
Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.

To solve it, we make the substitution

x =vy ...............................(2)

Differentiating equation (2) with respect to y, we get

(dx)/(dy) = v + y (dv)/(dy)

Substituting the value of x and (dx)/(dy) in equation (1), we get

v + y (dv)/(dy) = ( 2v e^v -1)/(2e^v)

or y(dv)/(dy) = ( 2v e^v -1)/(2e^v) -v

or y (dv)/(dy) = - 1/(2 e^v)

or 2e^v dv = (-dy)/y

or  int 2e^v * dv = - int (dy)/y

or  2 e^v = -log|y| +C

and replacing v by x/y, we get

2e^(x/y) + log |y| =C .....................(3)

Substituting x = 0 and y = 1 in equation (3), we get

2 e^0 + log |1| = C ⇒ C = 2

Substituting the value of C in equation (3), we get

2e^(x/y) + log |y| =2

which is the particular solution of the given differential equation.
Q 3117756689

Show that the family of curves for which the slope of the tangent at any point (x, y) on it is (x^ +y^2)/(2xy), is given by x^2 -y^2 =cx
Class 12 Chapter 9 Example 18
Solution:

We know that the slope of the tangent at any point on a curve is (dy)/(dx)

Therefore, (dy)/(dx) = (x^2 +y^2)/(2xy)

or (dy)/(dx) = (1 + y^2/x^2)/((2y)/x) ....................(1)

Clearly, (1) is a homogenous differential equation. To solve it we make substitution

y = vx

Differentiating y = vx with respect to x, we get

(dy)/(dx) = v + x (dv)/(dx)

or  v +x (dv)/(dx) = (1 +v^2)/(2v)

or x(dy)/(dx) = (1-v^2)/(2v)

(2v)/(1-v^2) dv = (dx)/x

or  (2v)/(v^2 -1) dv = -(dx)/x

Therefore int(2v)/(v^2 -1) dv = - int 1/x dx

or log | v^2 – 1 | = – log | x | + log |C_1 |

or log | (v^2 – 1) (x) | = log |C_1|

or (v^2 – 1) x = ± C_1

Replacing v by y/x , we get

(y^2/x^2 - 1)x = ± C_1

or (y^2 -x^2) pm C_1x  or  x^2 -y^2 =Cx`