Mathematics Homogeneous Equations and Linear First Order Differential Equation

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Homogeneous Equations and Linear First Order Differential Equation

Solution of Homogeneous Equations :

`text(Homogeneous Equations :)`
The function `f(x, y)` is said to be a homogeneous function of degree `n` if for any real number `t (ne 0)`, we have `f(tx, ty) = t^n (x, y)`. For example, `f(x, y) = ax^(2/3) + hx ^(1/3) xx y^(1/3) + by^(2/3)` is a homogeneous function of degree `2/3`.

`text(Solution of Homogeneous Differential Equation :)`
A differential equation of the form `dy/dx = (f(x,y)) / (phi (x,y))` , where `f(x, y)` and `f(x, y)` are homogeneous function of `x` and `y`, and of the same degree, is called Homogeneous. This equation may also be reduced to the form `dy/dx= g(x/y)` and is solved by putting `y = vx` so that the dependent variable `y` is changed to another variable ` v`, where `v` is some unknown function, the differential equation is transformed to an equation with variable separable.

`text(Equations Reducible to the Homogeneous Form :)`
Equation of the form `dy/dx = (ax+by +c) /(Ax +By +C) ` (`aB ne Ab` and `A+b ne 0`) can be reduced to a homogeneous form by changing the variable `x,y`, to `X, Y` by writing `x = X +h` and `y = Y + k`; where `h, k` are constant to be chosen so as to make the given equation homogeneous. We have

`dy /dx = (d (Y+k))/(d (X+h)) =(dY)/dX`

Hence the given equation becomes,

`(dY)/dX = (aX+ bY+ (ah+bk+c) )/(Ah + Bk + (Ah+Bk+C))`

Let `h` and `k` be chosen to satisfy the relation `ah + bk + c = 0` and `Ah + Bk + C = 0`.

`text(Homogeneous Equations :)`
The function `f(x, y)` is said to be a homogeneous function of degree `n` if for any real number `t (ne 0)`, we have `f(tx, ty) = t^n (x, y)`. For example, `f(x, y) = ax^(2/3) + hx ^(1/3) xx y^(1/3) + by^(2/3)` is a homogeneous function of degree `2/3`.

`text(Solution of Homogeneous Differential Equation :)`
A differential equation of the form `dy/dx = (f(x,y)) / (phi (x,y))` , where `f(x, y)` and `f(x, y)` are homogeneous function of `x` and `y`, and of the same degree, is called Homogeneous. This equation may also be reduced to the form `dy/dx= g(x/y)` and is solved by putting `y = vx` so that the dependent variable `y` is changed to another variable ` v`, where `v` is some unknown function, the differential equation is transformed to an equation with variable separable.

`text(Equations Reducible to the Homogeneous Form :)`
Equation of the form `dy/dx = (ax+by +c) /(Ax +By +C) ` (`aB ne Ab` and `A+b ne 0`) can be reduced to a homogeneous form by changing the variable `x,y`, to `X, Y` by writing `x = X +h` and `y = Y + k`; where `h, k` are constant to be chosen so as to make the given equation homogeneous. We have

`dy /dx = (d (Y+k))/(d (X+h)) =(dY)/dX`

Hence the given equation becomes,

`(dY)/dX = (aX+ bY+ (ah+bk+c) )/(Ah + Bk + (Ah+Bk+C))`

Let `h` and `k` be chosen to satisfy the relation `ah + bk + c = 0` and `Ah + Bk + C = 0`.

Linear Differential Equations :

A differential equation is said to be linear if the dependent variable & all its differential coefficients occur in degree one only and are never multiplied together. The nth order linear differential equation is of the form ;

`a_0 (x) (d^n y)/(dx^n) +a_1 (x) (d^(n-1)y)/(dx^(n-1)) +......+ a_n (x) * y = phi (x) ` . Where `a_0 (x) , a_1(x) ......a_n (x) `
are the coefficients of the differential equation.

`text(Solution of Linear Differential Equations of First Order :)`

The most general form of a linear differential equations of first order is `dy/dx+ Py=Q` ,
where `P` & `Q` are functions of `x` (Independent variable).

For solving such equations we multiply both sides by

Integrating factor =` I. F. = e^(int Pdx)`

So we get `e^(int Pdx) ((dy/dx )+py) = Q e^(int Pdx)`

`=> dy/dx e^(int Pdx) + y Pe ^(int Pdx) = Q e^(int Pdx)`

`=> d/dx (y e^(int Pdx))= Q e^(int Pdx)` [ since `d/dx (e^(int Pdx))=Pe^(int Pdx)` ]

`=> int d/dx (y e^(int Pdx)) dx= int Q(e^(intPdx))dx`

`=> y e^(int Pdx) = intQe^(int Pdx) +C`

which is the required solution of the given differential equation.

In some cases a linear differential equation may be of the form `dy/dx +P_1 x= Q_1`, where `P_1` and `Q_1` are
functions of `y` alone or constants. In such a case the integrating factor is `e^(int P_1dy)` , and

solutions is given by `x e^(int P_1 dy) = int Q_1 e^(int P_1 dy) dy +C`

`text (Equations Reducible To Linear Form` `text(Bernoulli's Equation):`

The equation `dy/dx +py =Q * y^n` where `P` & `Q` functions of `x` , is reducible to the linear form by
dividing it by `y^n` & then substituting `y^(-n+1) = Z`. Its solution can be obtained as in the normal case.

A differential equation is said to be linear if the dependent variable & all its differential coefficients occur in degree one only and are never multiplied together. The nth order linear differential equation is of the form ;

`a_0 (x) (d^n y)/(dx^n) +a_1 (x) (d^(n-1)y)/(dx^(n-1)) +......+ a_n (x) * y = phi (x) ` . Where `a_0 (x) , a_1(x) ......a_n (x) `
are the coefficients of the differential equation.

`text(Solution of Linear Differential Equations of First Order :)`

The most general form of a linear differential equations of first order is `dy/dx+ Py=Q` ,
where `P` & `Q` are functions of `x` (Independent variable).

For solving such equations we multiply both sides by

Integrating factor =` I. F. = e^(int Pdx)`

So we get `e^(int Pdx) ((dy/dx )+py) = Q e^(int Pdx)`

`=> dy/dx e^(int Pdx) + y Pe ^(int Pdx) = Q e^(int Pdx)`

`=> d/dx (y e^(int Pdx))= Q e^(int Pdx)` [ since `d/dx (e^(int Pdx))=Pe^(int Pdx)` ]

`=> int d/dx (y e^(int Pdx)) dx= int Q(e^(intPdx))dx`

`=> y e^(int Pdx) = intQe^(int Pdx) +C`

which is the required solution of the given differential equation.

In some cases a linear differential equation may be of the form `dy/dx +P_1 x= Q_1`, where `P_1` and `Q_1` are
functions of `y` alone or constants. In such a case the integrating factor is `e^(int P_1dy)` , and

solutions is given by `x e^(int P_1 dy) = int Q_1 e^(int P_1 dy) dy +C`

`text (Equations Reducible To Linear Form` `text(Bernoulli's Equation):`

The equation `dy/dx +py =Q * y^n` where `P` & `Q` functions of `x` , is reducible to the linear form by
dividing it by `y^n` & then substituting `y^(-n+1) = Z`. Its solution can be obtained as in the normal case.

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