Ma 221 Homework Solutions Due Date: January 24, 2012

Transcription

1 Ma Homewk Solutions Due Date: January, 0. pg. 3 #, 3, 6,, 5, 7 9,, 3;.3 p.5-55 #, 3, 5, 7, 0, 7, 9, (Underlined problems are handed in) In problems, and 5, determine whether the given differential equation is separable. ) y 3y y 3y therefe, this equation is separable. y 3y 3.) ds tlns t 8t ds t lns 8t ds t lns ds lns therefe, this equation is separable. t 6.) s ds s st Writing the equation in the fm ds s st shows that the equation is not separable. s In problems and 5, solve the equation. ) sec y x sec y x sec y x Using trigonemtric identities we have: secy sec y cos y x and cos y cosy cos y x cos y x y siny arctanx C y siny arctanx C y siny arctanx C 5.) y ye cos x sinx 0

2 Substituting: ye cos x sinx y e cos x sinx y let u cosx du sinx du sinx e u du y e u y C e cos x y C y C e cos x In problems 7, 9,,, 3, and 5, solve the initial value problem. 7) y x 3 y, y0 3 x 3 y y x 3 y x 3 ln y y exp C x Ce x Substituting the IC y0 3 3 Ce 0 C y e x Since y0 3 0, on an interval containing x 0 one has yx 0 and so yx yx. The solution is then: 9) y e x y e x y cosx, y 0 y cosx y / cosx y / cosx y / sinx C Substituting the IC y 0 0 / sin C C y / sinx y sinx sin x sinx

3 .) d ysin y, y y y sind Integrating The IC implies so y y y sind lny sin cos C C y lny sin cos ) x y 0, y0 x y x y x 3 3 y C Substituting the IC y0 : C C x 3 y y x ) t cos y, y0 cos y t sec y t tany t C tan C Thus C tany t 3

4 y arctant 5.) x y y0 3 x y x ln y y 3 x3 C y e 3 x3 C y Ce 3 x3 3 y0 C C y e 3 x3.3 p.5 to p. 55 #,, 6, 7, 9, 0,, 3, 5, 7, 8, 9,,, 30 (Underlined Problems are handed in) F problems, 3, and 5 determine whether the given equation is separable, linear, neither, both..) x sinx y Isolating we get: y sinx. Since the right hand side cannot be represented as a product gxpy,the x equation is not seperable..) t yt y This is a linear equation with independent variable t and dependent variable y.thisisa seperable equation as shown: Isolating yt t we get: y t t t t y gtpy 6.) x t x sint In this equation, the independent variable is t and the dependent variable is x. Dividing by x we obtain sint x t. Therefe, it is neither linear, because of the sint x term, n separable, because the righthand side is not a product of functions of single variables x and t. F problems 8, 9,, 3 and 5, obtain the general solution to the equation 8.) y e3x 0 In this equation, Px andqx e 3x The integrating fact is thus: x exp Px exp e x Multiplying both sides of the equation by x and integrating yields: e x e x y e x e 3x e x de x y e x

5 e x y e x ex C y ex Ce x e3x Ce x 9.) x y x 3 Putting the equation in standard fm: x y x Find x e P e x e ln x x x Multiply through by x to get x xy x d yx x d yx x yx x C y x 3 Cx.) t y 0 Choosing t as the independent variable and y as the dependent variable, the equation can be put into standard fm: t y 0 y t Thus:Pt andt exp e t Multiplying both sides by t and integrating yields: t e e t y t e t de t y t e t e t y t e t t e t e t t e t e t C t e t C y e t t e t C t Ce t 3.) y x 5y3 In this problem, the independent variable is y and the dependent variable is x. So, we divide the equation by y to rewrite it in standard fm. y x 5y3 y x 5y Therefe, Py y and the integrating fact, y, is y e y e ln y y y Multiplying the equation (in standard fm) by y and inegrating yield y x 5y y y yx 5y d y x 5y y x 5y y 5 C x y y 5 C y 3 Cy 5

6 5.) x xy x 0 Divide by x x y x x x x so Px x Find x e Px x e x x e lnx x / Multiply through by x to get x y x x y x d x y x x x Integrating gives yx / x / C y Cx / F problems 7, 8 and 9,, solve the initial value problems. 7.) y x xex y e This is a linear equation with Px /x and Qx xe x. The integrating fact is given by: x e P e x e lnx x ; F x 0 Multiply through by x to get x y y e x x x e x y x e x C y xe x Cx Plug in initial condition y e and solve f C y x e e C C Plug in the value f C y xe x x Using SNB to check our answer we have y x xe x y e, Exact solution is: xex x 8.) y e x 0;y0 3 6

7 y e x Find x e P e e x Multiply through by x to get e x y e x y e 3x d ex y e 3x ye x 3 e3x c y 3 e x C e x Plug in initial condition y e 0 C e 0 C 3 3 So C Plug in the value f C y 3 e x e x Using SNB to check our answer we have y e x 0 y0 3 and solve f C, Exact solution is: yx 3 e x e x 9.) t 3tx t lnt, x 0 In this problem, t is the dependent variable and x is the dependent variable. Dividing by t we have 3 t x t lnt t The integrating fact is e 3 t t 3. Thus t 3 3t x t 5 lnt t d t3 x t 5 lnt t Integrating we have since t 5 lnt 6 t6 lnt 36 t6 c The IC implies so c 7 36 and t 3 x 6 t6 lnt 36 t6 t c 0 36 c 7

8 t 3 x 6 t6 lnt 36 t6 t 7 36 x 6 t3 lnt 36 t3 t 7 36t 3.) cosx ysinx xcos x, y Putting the equation in standard fm: sinx cos x y xcosx 5 3 tanxy xcosx Find x e P e tan x e ln cos x cosx At the initial point,x,cos Multiply through by x to get cos x sinx y x d cos x d y cos x x y cos x x C y x cosx Ccosx Plug in initial condition y 5 cos Ccos 3 So C Plug in the value f C y x cosx cosx cosxx 0 therefe we can take x cosx y cos x x 5 3 and solve f C.) sinx ycosx xsinx, y Putting the equation in standard fm: cos x sinx y x Find x e P e cos x sinx cotxy x Multiply through by x to get sinx ycosx xsinx d d ysinx xsinx ysinx sinx xcosx C e lnsinx sinx ysinx xsinx (Using integration by parts) y x cos x C xcotx C sinx sinx sinx Plug in initial condition y and solve f C cos C sin sin So C Plug in the value f C y xcotx xcotx cscx sinx 8

9 30.) Show that the substitution v y 3 reduces equation y xy to the equation dv 6v 3x. Then solve the equation f v, and make the substitution v y3 to obtain the solution the equation y xy. y xy Divide by y y3 x v y 3 Differentiate v with respect to x dv 3y Divide by 3 y dv y 3 Notice that 3 dv is equal to the first term on the left hand side of the equation. Make that substitution. dv v x 3 Now multiply by 3 to get a first der linear differential equation. dv 6v 3x...(First der linear differential equation) Find x x e 6 e 6x Multiply through by x to get e 6x v 6e 6x v d e6x v 3xe 6x Now integrate ve 6x e 6x 3x xe6x e6x C Where SNB was used to evaluate the integral. Solve explicitly f v yields v x Ce 6x Plug y back into the equation from the substitution v y 3 and solve f y. y 3 x Ce 6x y x Ce 6x 3 9