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Misc 2 - Chapter 9 Class 12 Differential Equations - Part 7

Misc 2 - Chapter 9 Class 12 Differential Equations - Part 8


Misc 2 For each of the exercise given below , verify that the given function (π‘–π‘šπ‘π‘™π‘–π‘π‘–π‘‘ π‘œπ‘Ÿ 𝑒π‘₯𝑝𝑙𝑖𝑐𝑖𝑑) is a solution of the corresponding differential equation . (iii) 𝑦=π‘₯ sin⁑3π‘₯ : (𝑑^2 𝑦)/(𝑑π‘₯^2 )+9π‘¦βˆ’6 cos⁑〖3π‘₯=0γ€— 𝑦=π‘₯ sin⁑3π‘₯ Differentiating w.r.t x 𝑦^β€²=(π‘₯ 𝑠𝑖𝑛⁑3π‘₯ )^β€² 𝑦^β€²=π‘₯^β€² sin⁑3π‘₯+π‘₯(sin⁑3π‘₯)β€² 𝑦^β€²=sin⁑3π‘₯+π‘₯Γ—3 cos⁑3π‘₯ 𝑦^β€²=sin⁑3π‘₯+3π‘₯ cos⁑3π‘₯ Differentiating again w.r.t. x 𝑦^β€²β€²=(sin⁑3π‘₯ )^β€²+(3π‘₯ cos⁑3π‘₯ )^β€² 𝑦^β€²β€²=3 cos⁑3π‘₯+γ€–3(π‘₯)γ€—^β€² cos⁑3π‘₯+3π‘₯ (cos⁑3π‘₯ )^β€² 𝑦^β€²β€²=3 cos⁑3π‘₯+3 cos⁑3π‘₯+3π‘₯(βˆ’3 sin⁑3π‘₯) 𝑦^β€²β€²=6 cos⁑3π‘₯βˆ’9π‘₯ sin⁑3π‘₯ Putting 𝑦=π‘₯ sin⁑3π‘₯ 𝑦^β€²β€²=6 cos⁑3π‘₯βˆ’9𝑦 𝑦^β€²β€²+9π‘¦βˆ’6 cos⁑3π‘₯=0 Thus, Given Function is a solution of the Differential Equation

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.