MAP
2302-practice-final-soln

# The amplitude is radicalbig 6 5 the period is 2 π ω

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The amplitude is radicalbig 6 / 5 , the period is 2 π / ω = 2 π / 5 , and the phase angle is φ = arctan( - 5). a. Solve the initial value problem y = (1 + y 2 ) tan x with initial condition y (0) = 3. Solution: This equation is separable, so we have 1 1 + y 2 dy dx = tan x, integraldisplay 1 1 + y 2 dy dx dx = integraldisplay tan x dx, integraldisplay 1 1 + y 2 dy = integraldisplay tan x dx, arctan y = - ln(cos x ) + C. We can find C now by plugging in x = 0: arctan 3 = - ln(cos 0) + C (= C ) . So the final answer is y ( x ) = tan( - ln(cos x ) + arctan 3) . If we wanted to, we could further simplify this; arctan 3 = π / 3 , and tan is an odd function, so tan( - x ) = - tan x , so y ( x ) = - tan(ln(cos x ) - π / 3 ) . b. Use the substitution v = y / x to solve the differential equation dy dx = y 2 + 2 xy x 2 . Please remember to express your solution as a function of y .

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MAP 2302, Fall 2011 — Final Exam Review Problems 10 Solution: We are told to substitute v = y / x , but what do we substitute for dy / dx ? We find this by differentiation: v = y x , y = xv, dy dx = v + x dv dx (by the product rule) . Now we are ready to make the substitution: dy dx = y 2 + 2 xy x 2 , dy dx = parenleftBig y x parenrightBig 2 + 2 y x , v + x dv dx = v 2 + 2 v, x dv dx = v 2 + v, 1 v 2 + v dv dx = 1 x , integraldisplay 1 v 2 + v dv dx dx = integraldisplay 1 x dx, integraldisplay 1 v 2 + v dv = integraldisplay 1 x dx, The integral on the left is a bit tricky. We need to use partial fractions to integrate it: 1 v 2 + v = 1 v ( v + 1) = A v + B v + 1 , so by clearing denominators we see that we need 1 = A ( v + 1) + Bv, which gives us A = 1 and B = - 1. Therefore integraldisplay 1 v 2 + v dv = integraldisplay 1 v - 1 v + 1 dv = ln v - ln( v + 1) = ln parenleftbigg v v + 1 parenrightbigg .
MAP 2302, Fall 2011 — Final Exam Review Problems 11 Since the integral of 1 / x is ln x , we get that ln parenleftbigg v v + 1 parenrightbigg = ln x + C, v v + 1 = e ln x + C , v v + 1 “=” Cx, v = Cx 1 - Cx , y x = Cx 1 - Cx , y = Cx 2 1 - Cx . c. Use the substitution v = y 2 to solve the differential equation dy dx - y = e 2 x y 3 . Please remember to express your solution as a function of y . Solution: Even if we didn’t recognize this equation as a Bernoulli equation, we are given the substitution v = y 2 . In order to use this substitution, we divide the differential equation by the highest power of y to obtain y 3 dy dx - y 2 = e 2 x . Now we use the substitution: v = y 2 , dv dx = - 2 y 3 dy dx , which shows that y 3 dy dx = - 1 2 dv dx , so - 1 2 dv dx - v = e 2 x , or equivalently, dv dx + 2 v = - 2 e 2 x .

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MAP 2302, Fall 2011 — Final Exam Review Problems 12 This is a linear equation, with integrating factor μ = e integraltext 2 dx = e 2 x . Multiplying both sides by the integrating factor we get d dx ( μv ) = - 2 e 4 x , integraldisplay d dx ( μv ) dx = integraldisplay - 2 e 4 x dx, μv = - 1 2 e 4 x + C, so we have v = - 1 2 e 4 x + C μ = - 1 2 e 2 x + Ce 2 x . Finally, we replace v with y 2 and solve for y : 1 y 2 = - 1 2 e 2 x + Ce 2 x , y 2 = 1 - 1 2 e 2 x + Ce 2 x , y = ± radicalBigg 1 - 1 2 e 2 x + Ce 2 x .
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• Fall '08
• TUNCER
• Laplace, lim, Constant of integration, Exam Review Problems

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