he IVP d's (p² – 121) dp1 d's ds + s 1 = sin(p) dp3 p2 +1 dp s(7) = -994 ds = 19 dp l7 d's = 4 dp2 17 ds = 10 dp3 as a unique solution defined on the interval

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A: Second option is correct answer i. e.  I= -1-cosh(π)+π(i-1) . Thankyou !

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A: For the first question . Option first is correct answer . Thankyou !

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Q: 7. Obtain the solution of (D² – 4)y = 16 + cos 2x using the method of undetermined coefficients %3D

A: Since you have asked multiple question, we will solve the first question for you. If youwant any…

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A: A

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Q: 4. Verify that the solution of the DE d²x dt² + k²x = 0 is x = sin(kt).

A: To verify that the solution of the DE d2xdt2+k2x=0 is x=sinkt.

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A: Here I am attaching image so that you understand each and every step.

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