DIFFERENTIAL EQUATIONS

A Note Order and degree (if defined) of a differential equation are always positive integers. Example Find the order and degree, if defined, of each of the following differential equations: (i) dx − (ii) (iii) e ′ ′′′ + Solution (i) The highest order derivative present in the differential equation is dy dx , so its order is one. It is a polynomial equation in y ′ and the highest power raised to dy is one, so its degree is one. (ii) The highest order derivative present in the given differential equation is , so its order is two.

It is a polynomial equation in and dy dx and the highest power raised to is one, so its degree is one. (iii) The highest order derivative present in the differential equation is y ′′′ , so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined. EXERCISE .

Determine order and degree (if defined) of differential equations given in Exercises to . . sin( ′′′ . y ′ + y = .

ds d s s dt dt  + . . cos3 sin3 . y ′′′ + ( y ″ ) + ( y ′ ) + y = .

y ′′′ + y ″ + y ′ = . y ′ + y = e x . y ″ + ( y ′ ) + y = . y ″ + y ′ + sin y = .

The degree of the differential equation + = is (A) (B) (C) (D) not defined . The order of the differential equation is (A) (B) (C) (D) not defined . . General and Particular Solutions of a Differential Equation In earlier Classes, we have solved the equations of the type: x + = sin x – cos x = Solution of equations ( ) and ( ) are numbers, real or complex, that will satisfy the given equation i.e., when that number is substituted for the unknown x in the given equation, L.H.S.

becomes equal to the R.H.S.. Now consider the differential equation ... ( ) In contrast to the first two equations, the solution of this differential equation is a function φ that will satisfy it i.e., when the function φ is substituted for the unknown y (dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S..

The curve y = φ ( x ) is called the solution curve (integral curve) of the given differential equation. Consider the function given by y = φ ( x ) = a sin ( x + b ), ... ( ) where a , b ∈ R . When this function and its derivative are substituted in equation ( ), L.H.S.

= R.H.S.. So it is a solution of the differential equation ( ). Let a and b be given some particular values say a = and b π , then we get a function y = φ ( x ) = 2sin π ... ( ) When this function and its derivative are substituted in equation ( ) again L.H.S.

= R.H.S.. Therefore φ is also a solution of equation ( ). Function φ consists of two arbitrary constants (parameters) a , b and it is called general solution of the given differential equation. Whereas function φ contains no arbitrary constants but only the particular values of the parameters a and b and hence is called a particular solution of the given differential equation.

The solution which contains arbitrary constants is called the general solution ( primitive ) of the differential equation. The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation. Example Verify that the function y = e – x is a solution of the differential equation Solution Given function is y = e – x . Differentiating both sides of equation with respect to x , we get = − Now, differentiating ( ) with respect to x , we have = e – x Substituting the values of , d y dy and y in the given differential equation, we get L.H.S.

= e – x + (– e – x ) – . e – x = e – x – e – x = = R.H.S.. Therefore, the given function is a solution of the given differential equation. Example Verify that the function y = a cos x + b sin x , where, a , b ∈ R is a solution of the differential equation Solution The given function is y = a cos x + b sin x Differentiating both sides of equation ( ) with respect to x , successively, we get dx = – a sin x + b cos x = – a cos x – b sin x Substituting the values of and y in the given differential equation, we get L.H.S.

= (– a cos x – b sin x ) + ( a cos x + b sin x ) = = R.H.S. Therefore, the given function is a solution of the given differential equation. EXERCISE . In each of the Exercises to verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: .

y = e x + : y ″ – y ′ = . y = x + x + C : y ′ – x – = . y = cos x + C : y ′ + sin x = . y = : y ′ = .

y = A x : xy ′ = y ( x ≠ ) . y = x sin x : xy ′ = y + x ( x ≠ and x > y or x < – y ) . xy = log y + C : y ′ = ( xy ≠ ) . y – cos y = x : ( y sin y + cos y + x ) y ′ = y .

x + y = tan – y : y y ′ + y + = . y = a x ∈ (– a , a ) : x + y dy dx = ( y ≠ ) . The number of arbitrary constants in the general solution of a differential equation of fourth order are: (A) (B) (C) (D) . The number of arbitrary constants in the particular solution of a differential equation of third order are: (A) (B) (C) (D) .

. Methods of Solving First Order, First Degree Differential Equations In this section we shall discuss three methods of solving first order first degree differential equations. . .

Differential equations with variables separable A first order-first degree differential equation is of the form dx = F( x , y ) If F( x , y ) can be expressed as a product g ( x ) h ( y ), where, g ( x ) is a function of x and h ( y ) is a function of y , then the differential equation ( ) is said to be of variable separable type. The differential equation ( ) then has the form dx = h ( y ) . g ( x ) If h ( y ) ≠ , separating the variables, ( ) can be rewritten as ( ) h y dy = g ( x ) dx ... ( ) Integrating both sides of ( ), we get ( ) dy h y ( ) g x dx ...

( ) Thus, ( ) provides the solutions of given differential equation in the form H( y ) = G( x ) + C Here, H ( y ) and G ( x ) are the anti derivatives of ( ) h y and g ( x ) respectively and C is the arbitrary constant. Example Find the general solution of the differential equation , ( y ≠ ) Solution We have Separating the variables in equation ( ), we get ( – y ) dy = ( x + ) dx Integrating both sides of equation ( ), we get ( y dy ) y − x + y + x – y + C = x + y + x – y + C = , where C = 2C which is the general solution of equation ( ). Example Find the general solution of the differential equation Solution Since + y ≠ , therefore separating the variables, the given differential equation can be written as Integrating both sides of equation ( ), we get tan – y = tan – x + C which is the general solution of equation ( ). Example Find the particular solution of the differential equation dx = − given that y = , when x = .

Solution If y ≠ , the given differential equation can be written as y = – x dx Integrating both sides of equation ( ), we get