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3.1 Integrating Both Sides Implicit differentiation is a technique from calculus for taking the derivative of expressions involving an unknown function y ( x ) . The idea is to use the normal algorithm for differentiation, and to simply write y 0 whenever we would usually need to take the derivative of y. For example, to take the derivative of something like (sin x ) 3 , we would normally use the power rule and the chain rule: g d f (sin x ) 3 3 (sin x ) 2 cos x. dx Here the cos x comes from the chain rule, since cos x is the derivative of sin x. We use exactly the same procedure to take the derivative of something like y 3 , except that we write y 0 whenever we need the derivative of y: d f 3g y 3y 2 y 0 . dx Here the 3y 2 comes from the product rule, and we multiply by y 0 because of the chain rule. EXAMPLE 1 Take the derivative of both sides of the following equation: sin ( y ) + x 2 y x 3 . SOLUTION We can use the chain rule to take the derivative of sin ( y ) : g d f sin ( y ) cos ( y ) y 0 . dx The derivative of x 2 y requires the product rule: d f 2 g x y 2x y + x 2 y 0 . dx Thus the derivative is y 0 cos ( y ) + 2x y + x 2 y 0 3x 2 . In general, the product rule states that d f g uv u 0 v + uv 0 . dx In this case, u x 2 and v y, so u 0 2x and v 0 y 0 . We can use implicit differentiation to find a differential equation that has a given general solution. For example, suppose we want to find a differential equation whose general solution is y Cx 2 . We start by solving for the constant C: yx −2 C. If we now take the derivative of both sides, the C disappears: y 0 x −2 − 2yx −3 0. This is a differential equation whose general solution is y Cx 2 . We can simplify the equation a bit by multiplying through by x 3 : x y 0 − 2y 0. Since C is a constant, its derivative is zero. INTEGRATING BOTH SIDES 2 EXAMPLE 2 Find a differential equation whose general solution is y ln ( x + C ) . SOLUTION To solve this equation for C, we first take the exponential of both sides: e y x + C. Thus e y − x C. Taking the derivative of both sides gives e y y 0 − 1 0. This is the differential equation we wanted. We can simplify it by solving for y 0 : y 0 e −y . Integrating Differential Equations Now that we know how take derivatives implicitly, we can reverse the process to solve differential equations. For example, given a differential equation like 2y y 0 4x 3 + 2, Technically, there might be a constant on both sides of the antiderivative, i.e. we can integrate both sides to get an equation involving x and y: y 2 x 4 + 2x + C. y 2 + C1 x 4 + 2x + C2 However, we can combine the two constants by subtracting C1 from both sides of this equation. We can now solve for y to get the general solution to the given differential equation: p y ± x 4 + 2x + C. EXAMPLE 3 Find the general solution to the equation y 2 y 0 e 3x . SOLUTION We can integrate both sides to get 1 1 3 y e 3x + C. 3 3 Solving for y gives y p3 e 3x + 3C. Since C is an arbitrary constant, 3C is actually itself an arbitrary constant. If we let A 3C, we can rewrite our general solution as y where A is an arbitrary constant. p3 e 3x + A, INTEGRATING BOTH SIDES 3 EXAMPLE 4 Solve the following initial value problem: e 4y y 0 x, SOLUTION y (0) 1. In general, the antiderivative of f ( y ) y 0 is Integrating both sides yields Z 1 4y 1 e x 2 + C. 4 2 Though we could solve for y at this point, it is easier to start by substituting in the initial condition: 1 4 e 0 + C. 4 1 Then C e 4 , so the solution becomes 4 f ( y ) dy. In this case, the antiderivative of e 4y y 0 is Z e 4y dy 1 4y e + C. 4 1 1 1 4y e x2 + e 4 . 4 2 4 Solving for y gives y 1 2 ln 2x + e 4 . 4 Sometimes it is possible to recognize one side of an equation as the result of the product rule. EXAMPLE 5 Find the general solution to the differential equation x 3 y 0 + 3xx 2 y 1. The left side of this equation is the result of a product rule, with the two factors being x 3 and y. Integrating both sides gives SOLUTION x 3 y x + C. Solving for y yields the general solution: y x −2 + Cx −3 . A Closer Look Implicit Solutions Sometimes after you integrate both sides of a differential equation it is not possible to algebraically solve for y. For example, consider the equation 3y 2 + e y y 0 x 2 . Integrating both sides gives 1 3 x + C. 3 Unfortunately, there is no way to solve this equation algebraically for y. This implicit solution still describes the solutions of the equation, in the sense that the corresponding curves in the x y-plane are graphs of the solution functions (see Figure 1), but there is no way to write explicit algebraic formulas for these solutions. y3 + e y a Figure 1: Three curves of the form y3 + e y 1 3 x + C. 3 INTEGRATING BOTH SIDES 4 A Closer Look Integrating Second-Order Equations The method of integrating both sides can also be used on second-order equations, although the antiderivatives are often much more difficult. For example, consider the equation y y 00 + ( y 0 ) 2 6x. Though it may not be obvious, the left side is a result of the product rule: it is the derivative of the product y y 0 . Integrating both sides gives y y 0 3x 2 + C. We have now reduced to a first-order equation, and we can integrate both sides again to obtain the general solution. Specifically, we get 1 2 y x 3 + Cx + D, 2 where D is an arbitrary constant. Multiplying by two and taking the square root gives p y ± 2x 3 + Ax + B, where A 2C and B 2D are arbitrary constants. EXERCISES Take the derivative of both sides of the given equation with respect to x. 1–2 1. x y + ln y sin (2x ) 2. e 2y − x y 2 1 Find a differential equation whose general solution is the given formula. 3–6 √ √ 3. y C x 4. y ± x + C 5. y tan ( x + C ) 6. y e 2x + Ce −x Find the general solution to the given differential equation by integrating both 7–12 sides. 7. y y 0 e x √ 9. y 0 y cos x 11. e 3x y 0 + 3e 3x y e 5x 13–14 8. y 0 e 2y x 3 10. x 4 y 0 + 4x 3 y x 4 12. y 0 tan x + y sec2 x cos x Solve the given initial value problem. 13. y 0 cos y e −x , y (0) 0 14. −y −2 y 0 2x, y (0) 3