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5, Differential Equations, EQUATIONS OF THE FIRST ORDER AND, HIGHER DEGREE, In the lower classes, the students have studied differential equations of the first, order and first degree, such as variable separable equations and linear equations., Now we shall study differential equations of the first order and degree greater than, or equal to two., The general form of the differential equation of the first order and nh degree is, п-1, n-2, dy, + fi (x, y), dy, + f2 (x, y)|, dy, +..., dx, dx, dx, dy, + fn-1 (x, y)+ fn (x, y) = 0., dx, dy, by p for convenience, the general equation becomes, dx, If we denote, p" +fi (x, y) p"-1 +f2 (x, y) p"-2 + .. .+ fn-1 (x, y) p + f„(x, y) = 0, Since (1) is an equation of the first order, its general solution will contain only, one arbitrary constant. Solution of (1) will depend on solving one or more equations, of the first order and first degree., To solve (1), it is to be identified as an equation of any one of the following, types and then solved as per the procedure indicated below:, (1) Equations solvable for p., (ii) Equations solvable for y., (iii) Equations solvable for x., (iv) Clairaut's equations., (1), %3!, Scanned by CamScanner

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258 Engineering Mathematics, Type 1-Equations solvable for p, requation (1) is of this type, then the L.H.S. of (1) can be resolved inton linear factors., Then (1) becomes, (P- F) (p - F,)... (p- F) = 0, from which we get p F,p= F, where F,, F2,, .,F, are functions of x and y., Each of thesen equations is of the first order and first degree and can be solved, by methods already known., Let the solutions of the above n component equations be ø1 (x, y, c), $2 (x, y, c) = 0, ..., ¢, (x, y, c) = 0. Then the general solution of (1) is got by, combining the above solutions and given as ø (x, y, c) Ø (x, y, c)... , (x, y, c) = 0., Type 2-Equations solvable for y, If the given differential equation is of this type, theny can be expressed explicitly as a, single valued function of x and p., i.e. the equation of this type can be re-written as, y=f(x, p)., (1), Differentiating (1) with respect to x, we get, dp, x, p,, dx, p =, (2), Equation (2) is a differential equation of the first order and first degree in the variables, x and p. It can be solved by methods already known. Let the solution of (2) be w(x,, p, c) = 0... (3), where c is an arbitrary constant. If we eliminate p between (1) and (3),, the eliminant is the general solution of the given equation., If p cannnot be easily eliminated between (1) and (3), they jointly provide the, required solution in terms of the parameter p., Type 3-Equations solvable for x, If the given differential equation is of this type, thenx can be expressed explicitly as a, single valued function of y and p., i.e the equation of this type can be re-written as, x=f(y, p), Differentiating (1) with respect to y, we get, (1), у, р,, Equation (2) is a differential equation of the first order and first degree in the, variables y and p. It can be solved by methods already known., Let the solution of (2) be, y(y, p, c) = 0, a (3), where c is an arbitrary constant., Scanned by CamScanner

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zÁDifferential Equations 259, If we climinate p betwcen (1) and (3), the eliminant is the general solution of the, given equation., If p cannot be casily eliminated between (1) and (3), they jointly provide the, required solution in terms of the parameter p., Note: Some differential equations can be put in both the forms y f(x, p) and, x= f(y, p). Both these forms may lead to the required solution. Sometimes one of, the forms will lead to the required solution more easily than the other., Type 4-Clairaut's equations, An cquation of the formy = px +f(p) is called Clairaut's equation., Clairaut's equation is only a particular case of type-2 equation. Hence it can be, solved in the same way in which a type-2 equation is solved, as explained below:, Let the Clairaut's equation be, y = px + f(p), (1), Differentiating (1) w. r. t. x,, dp, p =p+ {x+f° (p)}, dx, dp, = 0... (2) or f'(p)+ x=0', :., (3), dx, (4), Solving (2), we get p = c, Eliminating p between (1) and (4), we get the general solutiíon of (1) as, y = cx +f(c)., Thus the general solution of a Clairaut's equation is obtained by replacing p by, c in the given equation., Eliminating p between (1) and (3), we also get a solution of (1)., This solution does not contain any arbitrary constant. Also it connot be obtained, as a particular case of the general solution. This solution is called the singular, solution of the equation (1)., Note: The singular solution of (1) is the eliminant of p between y = px + f(p) and, df, = 0. Equivalently, the singular solution of (1) is the eliminant of c between y, dp, = cx + f (c) and x +, observe that if the general solution of (1), i.e. y = cx + f (c) represents a family of, straight lines, then the singular solution represents the envelope of the family of, straight lines., df, = 0. Hence from the topic 'Envelopes' in Chapter 3, we, de, Worked Example 5(a), 2, dy, 8., + 15 = 0., Example I Solve the equation, dr, dr, The given equation is p2-8p + 15= 0, which is solvable for p., The equation is (p- 3) (p- 5) = 0, dy, = 3, dx, dy, = 5, dx, or, Scanned by CamScanner