`color{red} ♦ ` Solution of 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)`

`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

Class 12 Chapter 9 Example 15

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

Class 12 Chapter 9 Example 16

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

Class 12 Chapter 9 Example 17

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

Class 12 Chapter 9 Example 18

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_1`x ` or ` x^2 -y^2 =Cx`