Worksheet 9. Math 1B, GSI: Andrew Hanlon. 1 Ce 3t 1/3 1 = Ce 3t. 4 Ce 3t 1/ =

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1 Worksheet 9 Math B, GSI: Andrew Hanlon. Show that for each of the following pairs of differential equations and functions that the function is a solution of a differential equation. (a) y 2 y + y 2 ; Ce 3t /3 where C is a constant. (b) y + y 0; A cos t + B sin t where A, B are constants. (c) y (n) + a y (n ) a n y + a n y 0; e rt where r is a root of the polynomial x n + a x n a n x + a n. Solution: In each case, we simply plug in. (a) We first compute ( d dt and ( Then, we see (b) We hae ) Ce 3t /3 ) Ce 3t / ( ) ( + Ce 3t /3 Ce 3t (Ce 3t /3) 2 4 Ce 3t /3 + (Ce 3t /3) 2. ) 2 Ce 3t /3 Ce 3t / Ce 3t /3 + (Ce 3t /3) 2 3 Ce 3t /3 + (Ce 3t /3) 3(Ce 3t /3) + 2 (Ce 3t /3) 2 3Ce 3t (Ce 3t /3) 2 ( Ce 3t /3 (A cos t + B sin t) + A cos t + B sin t ( A sin t + B cos t) + A cos t + B sin t ) A cos t B sin t + A cos t + B sin t 0 (c) Note that that (e rt ) (k) dk dx k (e rt ) r k e rt. Also, since r is a root of the polynomial, we know r n + a r n a n r + a n 0.

2 Thus, we hae (e rt ) (n) + a (e rt ) (n ) a n (e rt ) + a n e rt r n e rt + a r n e rt a n re rt + a n (r n + a r n a n r + a n )e rt 0e rt 0 2. One method of drawing of direction fields is called the isocline method. To do this, you set the deriatie to a constant alue to obtain an equation for the points in the plane where the deriatie takes that alue and then draw the graph of this equation. Use this method or any other method to draw the direction fields of the following differential equations. (a) y x y (b) y x y 2 Solution: (a) Note that in this case the isoclines are straight lines. (b) In this case, the isoclines are circles. 3. Sole the following initial alue problems. (a) y x + xy y, y(0) 0 (b) xy 2y x 2, y() 0 Solution: (a) We hae dx x + xy y x + y(x ) (y + )(x ) 2

3 which can be rearranged to Integrating both sides gies y + (x ) dx. ln y + x2 2 x + C which rearranges (after altering C and remoing the unnecessary absolute alue) to Then, applying the initial condition gies so C. Therefore, is the solution to the initial alue problem. (b) We rewrite the equation as y Ce x2 /2 x. 0 y(0) Ce 0 C y e x2 2 x y 2 x y x which is in standard linear form. Thus, our integrating factor is Multiplying my the integrating factor gies Then, integrating gies and thus µ(x) e 2/x dx e 2 ln x x 2. x 2 y 2 x y d 2 dx Now, applying the initial condition gies Therefore, is the solution to the initial alue problem. y ln x + C x2 ( ) x y 2 x. y x 2 ln x + Cx 2. 0 y() 0 + C C. y x 2 ln x 4. Find a family of orthogonal cures to the family of circles gien by x y 2 2cx. Hint: to sole the differential equation that you obtain, you will need to substitute y/x. Solution: We differentiate to get 2x + 2y dx 3 2c x + y2 x

4 where we remoed the dependence on c. This rearranges to dx y2 x 2 2yx. Now to get an orthogonal family, we sole a differential equation so the that the tangents are the negatie reciprocal as follows. dx 2yx x 2 y. 2 Now, we substitute y/x, that is, y x and /dx xd/dx + to get which we rearrange to x d dx + 2x2 x 2 x x d dx or 2 + d 3 x dx Now, we wish to take the integral of both sides. Note that by doing a partial fractions decomposition so 2 + d 3 d 2 d ln u where we used a u-substitution u. Then, we hae ln ln x + C so we get (after modifying C) that Substituting back, we get that is Cx Cx. y/x (y/x) y x 2 Cy after modifying C which is another family of circles. yx y x 2 du ln ln u ln 4

5 5. A tank initially contains 0 kg of salt and 00 L of water completely mixed. Pure water flows in at 0 L/min and the contents of the tank flow out at 0 L /min. Salt is also added to the tank at 0. kg/min. How much salt is in the tank after hour? About how much salt is in tank after a ery long time (t )? Solution: Let S denote the amount of salt in kg. Since the olume is constant 00L, we get ds S salt in salt out 0. dt 00 0 ( S) 0 which has units kg/min. We rearrange this, to get We integrate to get and we get ds S dt 0. ln S t 0 + C S Ce t/0 +. Since the initial amount of salt is 0 kg, we hae so C 9. The salt at hour is 0 S(0) C + S(60) 9e 6 + kg.022 kg. We also hae lim S(t) kg. t (note that at this point, the salt in and out is equal) 6. Use Euler s method to estimate the alue of the solution of y e t2, y(0) 0 at t with a step size of 0.. Solution: We compute y(0.) y (0)(0. 0) + y(0) e 0 (0.) y(0.2) y (0.)(0.2 0.) + y(0.) e (0.)2 (0.) y(0.3) y (0.2)( ) + y(0.2) e (0.2)2 (0.) y(0.4) y (0.3)( ) + y(0.3) e (0.3)2 (0.) y(0.5) y (0.4)( ) + y(0.4) e (0.4)2 (0.) y(0.6) y (0.5)( ) + y(0.5) e (0.5)2 (0.) y(0.7) y (0.6)( ) + y(0.6) e (0.6)2 (0.) y(0.8) y (0.7)( ) + y(0.7) e (0.7)2 (0.) y(0.9) y (0.8)( ) + y(0.8) e (0.8)2 (0.) y() y (0.9)( 0.9) + y(0.9) e (0.9)2 (0.) Note that in this case we are simply estimating 0 e t2 dt and this integral cannot be soled in terms of simple functions. 5