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Q1:Write the second order inhomogeneous linear differential equationSee Answer
Q2:16. sin(xy) = cos(x + y) 17. √√x+y=x² + y²¹ Answer + 18. sin x cos y = x² — 5y 19. √√xy=1+x²y Answer 20. xy = √x² + y²See Answer
Q3: 5) Solve the Non-linear second ODE: 3 x^{2} y \frac{d^{2} y}{d x^{2}}+y^{2}=x^{2}\left(\frac{d y}{d x}\right)^{2}See Answer
Q4: Determine the differentials with respect to x of the following. Take care to show your method clearly and to simplify the results.1. \text { a) } \sin (x) \cdot \cos (x) \text { b) } \ln \left(3 x^{2}\right) \text { c) } x \sqrt{2 x^{2}+14} \text { d) } \frac{e^{x}-e^{-x}}{2} \text { e) } \tan (x) \text { f) } 6 a x^{3}-4 a cSee Answer
Q5: Find the value of x at the turning points on the curve of y=x^{3}-6 x^{2}+9 x-2See Answer
Q6: 4) Solve the ODE by using Superposition Method: y^{\prime \prime}+3 y=x+3 e^{-3 x}See Answer
Q7: • Perform the indicated operation. Write the answer in lowest terms. \frac{a^{2}-5 a+4}{a^{2}-a-12} \div \frac{a-4}{a^{2}+5 a+6}See Answer
Q8: \text { 14. } \quad y^{\prime}+2 y=4 \sin (3 t), \quad y(0)=3See Answer
Q9: \text { 1. } y^{\prime \prime}+2 y^{\prime}+2 y=0, \quad y(0)=1, \quad y^{\prime}(0)=0See Answer
Q10: \text { 11. } Y(s)=\frac{2}{2 s+5}-\frac{3 s}{s^{2}+2}+\frac{1}{5 s^{2}}See Answer
Q11: Use Laplace transforms to solve each of the differential equations 5 - 8, in terms of the initial conditions y(0) and y'(0). Compare your solution to.the general solution you would obtain using the methods of Chapter 3 (identify the constants C1 and C2). \text { 6. } y^{\prime \prime}+2 y^{\prime}+5 y=0See Answer
Q12: Calculate the maximum volume of an open-toped box that can be made from a sheet of cardboard 47 cm long by 38 cm wide. See Answer
Q13: 3. If the scalar function u(x, y, z) has continuous derivatives, and the vector differential operator \nabla=\hat{\mathbf{i}} \frac{\partial}{\partial x}+\hat{\mathbf{j}} \frac{\partial}{\partial y}+\hat{\mathbf{k}} \frac{\partial}{\partial z} then prove the vector identity \nabla \times \nabla \psi=0See Answer
Q14: we iave denned tangent pianes for suraces z7x, y), where has contindousfirst partial derivatives.See Answer
Q15: 22. PROJECT. Further Results by Differentiation.Proceeding as in Example 1, obtain \text { (a) } \mathscr{I}(t \cos \omega t)=\frac{s^{2}-\omega^{2}}{\left(s^{2}+\omega^{2}\right)^{2}} and from this and Example 1: (b) formula 21, (c) 22,(d) 23 in Sec. 6.9. \text { (e) } \mathscr{L}(t \cosh a t)=\frac{s^{2}+a^{2}}{\left(s^{2}-a^{2}\right)^{2}} \text { (f) } \mathscr{L}(t \sinh a t)=\frac{2 a s}{\left(s^{2}-a^{2}\right)^{2}}See Answer
Q16: 23-29 INVERSE TRANSFORMS BY INTEGRATION \text { 23. } \frac{3}{s^{2}+s / 4} \text { 24. } \frac{20}{s^{3}-2 \pi s^{2}} \text { 25. } \frac{1}{s\left(s^{2}+\omega^{2}\right)} \text { 26. } \frac{1}{s^{4}-s^{2}} \text { 27. } \frac{s+1}{s^{4}+9 s^{2}} \text { 28. } \frac{3 s+4}{s^{4}+k^{2} s^{2}} \text { 29. } \frac{1}{s^{3}+a s^{2}}See Answer
Q17: 0. PROJECT. Comments on Sec. 6.2. (a) Give reasonswhy Theorems 1 and 2 are more important thanTheorem 3. (b) Extend Theorem 1 by showing that if f(o is continuous, except for an ordinary discontinuity (finite jump) at some t = a(>0), the other conditions remaining gas in Theorem 1. then (see Fig i17) \left(1^{*}\right) \mathscr{L}\left(f^{\prime}\right)=s \mathscr{S}(f)-f(0)-[f(a+0)-f(a-0)] e^{-a s} \text { (c) Verify }\left(1^{*}\right) \text { for } f(t)=e^{-t} \text { if } 0<t<1 \text { and } 0 \text { if } (d) Compare the Laplace transform of solving ODES with the method in Chap. 2. Give examples of your own to illustrate the advantages of the present method(to the extent we have seen them so far). See Answer
Q18: 12-15SHIFTED DATA PROBLEMS 12. у" - 2у - 3у =0, y(4) =-3 -3,У(4)-17 13. у - 6y= 0 y(-1)=4 14. y" + 2y' +5y = 50t - 100, y(2) = -4.y (2) = 14 15. y" + 3y – 4y = 6e-2t -3 y (1.5) = 4, y(1.5) = 5See Answer
Q19: 24. Changing temperature along a space curve The Celsius temperature in a region in space is given by T(x, y, z) = 2x² – xyz . A particle is moving in this region and its position at time t is given by x = 2t², y = 3t, z = -t², where time is measured in seconds and distance in meters.See Answer
Q20: 31-32 Show that the lines L, and L2 are skew. 31. L1: x= 1+7t, y = 3+t, z= 5-3t L2: x = 4 - t, y = 6, z = 7+2tSee Answer

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