Introduction to Differential Equations - Self Quiz Solutions 30 | MATH 220, Study notes of Differential Equations

Download Introduction to Differential Equations - Self Quiz Solutions 30 | MATH 220 and more Study notes Differential Equations in PDF only on Docsity! MATH 220 Self-quiz 30 1. Find the general solution of the differential equation: yy′ = sin( √ x) + sin( √ x)y4 Solution: This is a separable equation, so we begin by separating the equa- tion y dy dx = sin( √ x)(1 + y4) y 1 + y4 dy = sin( √ x)dx Then, we integrate both sides ∫ y 1 + y4 dy = ∫ sin( √ x)dx In order to integrate the left side, we will use substitution, u = y2 and du = 2ydy ∫ y 1 + y4 dy = 1 2 ∫ du 1 + u2 = 1 2 arctan(y2) + C In order to integrate the right side, we will use substitution, u = √ x and du = 1 2 √ x dx. We can see that du = 1 2u∫ sin( √ x)dx = 2 ∫ u sin(u)du = 2[−u cos(u)− ∫ cos(u)du] = = 2[−u cos(u)− sin(u)] + C = 2[− √ xcos( √ x)− sin( √ x)] + C Thus the solution is: 1 2 arctan(y2) = 2[− √ x cos( √ x)− sin( √ x)] + C 2. Find the general solution of the differential equation: ty′′ + (−t− 6)y′ + 6y = 0 given that y(t) = et satisfies the equation. Solution: We will use the method of variation of parameters. We are given y1, so we know that y2 has to be a function, u, multiplied by our y1. We have to find the value of y2. In order to do this, we substitute y2, y′2, and y′′2 in the original equation and solve for u. t(u′′et + 2u′et + uet) + (−t− 6)(u′et + uet) + 6uet = 0 tetu′′ + (tet − 6et)u′ = 0 ⇒ tu′′ + (t− 6)u′ = 0 In order to solve for u, we first need to set v = u′, solve for the value of v by using separable equations method, and integrate whatever we get for v (in this case we will see we need to apply integration by parts several times) to find u. t dV dt = (6− t)V ⇒ ∫ 1 V dV = ∫ ( 6 t − 1)dt lnV = 6lnt− t V = e6lnt · e−t = t6e−t u = ∫ t6e−tdt u = (−t6 − 6t5 − 30t4 − 120t3 − 360t2 − 720t− 720)e−t Now that we have a value for u, we can find what y2 is. 2