Differential Equations with Variables Separable

Solution of differential equations :  If we have a differential equation of order ‘ n ‘ then by solving a differential equation we mean to get a family of curves with n parameters whose differential equation is the given differential equation.

For the solution of differential equations of order 1 and degree 1 , we have the following methods:

First Order Differential Equations with Variables Separable :

Let the differential equation be of the form

$ \displaystyle \frac{dy}{dx} = f (x , y) $ . . . . (1)

where f(x, y) denotes a function of x and y.

(1) is separable if f(x , y) can be expressed in the form :

$ \displaystyle \frac{M(x)}{N(y)} \; or \; M(x).N(y) $

M(x), N(y) are real valued functions of x and y respectively.

Then , we have

$ \displaystyle \frac{dy}{dx} = \frac{M(x)}{N(y)} \; or \; M(x).N(y) $

or M(x) . N(y)

⇒ N(y) dy = M(x) dx or

$ \displaystyle \frac{dy}{N(y)} = M(x).dx $ . . . . (2)

Integrating both sides of (2), we get the solution viz.

$ \displaystyle \int \frac{dy}{N(y)} = \int M(x).dx + c $

Illustration : Solve the differential equation

$ \displaystyle x y \frac{dy}{dx} = \frac{1+y^2}{1+x^2}(1+x+x^2) $

Solution: Differential equation can be rewritten as

$ \displaystyle x y \frac{dy}{dx} = (1+y^2)(1+\frac{x}{1+x^2}) $

$ \displaystyle \frac{y}{1+y^2} \frac{dy}{dx} = \frac{1}{x}+\frac{1}{1+x^2} $

Integrating, we get

$\displaystyle \frac{1}{2}log(1+y^2)= logx + tan^{-1}x + logc $

$\displaystyle \sqrt{1+y^2} = c x e^{tan^{-1}x} $

Differential Equations Reducible to the Separable Variable Type:

Sometimes differential equation of the first order cannot be solved directly by variable separation. By some substitution we can reduce it to a differential equation with separable variables. A differential equation of the form dy/dx = f(ax + by + c) is solved by writing ax + by + c = t.

Illustration : Solve

$ \displaystyle \frac{dy}{dx} = sin^2(x+3y) + 5 $

Solution: Let x + 3y = t, so that

1 + 3(dy/dx) = dt/dx

$ \displaystyle \frac{dy}{dx} = \frac{1}{3}(\frac{dt}{dx}-1) $

The given differential equation becomes

$ \displaystyle \frac{1}{3}(\frac{dt}{dx}-1) = sin^2 t + 5 $

$ \displaystyle \frac{dt}{dx} = 3 sin^2 t + 16 $

$ \displaystyle \int \frac{dt}{3 sin^2 t + 16} = \int dx $

$ \displaystyle \int \frac{sec^2 t dt}{19 tan^2 t + 16} = \int dx $

Let , tant = u => sec2t dt = du

$ \displaystyle \int \frac{du}{19 u^2 + 16} = x $

$ \displaystyle x = \frac{1}{19} \int \frac{du}{u^2 + 16/19} $

$ \displaystyle x = \frac{1}{19}\frac{\sqrt{19}}{4}tan^{-1}\frac{\sqrt{19}u}{4} + c $

[ where u=tan t =tan( x+ 3y)]

Exercise :

Solve the following differential equations.

(i) $ \displaystyle sin^{-1}\frac{dy}{dx} = x + y $

(ii) $ \displaystyle \frac{dy}{dx} = e^{x-y} + x^2 e^{-y} $

(iii) $ \displaystyle \sqrt{1+x^2 +y^2 +x^2 y^2 } + x y \frac{dy}{dx} = 0 $

(iv) $ \displaystyle (\frac{x+y-a}{x+y-b})\frac{dy}{dx}= \frac{x + y +a}{x + y + b} $

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