Example 3 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
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Last updated at April 16, 2024 by Teachoo
Example 3 Verify that the function π¦=π cosβ‘γπ₯+π sinβ‘γπ₯, γ γ where , π, πβπ is a solution of the differential equation (π^2 π¦)/(ππ₯^2 )+π¦=0 π¦=π cosβ‘γπ₯+π sinβ‘γπ₯ γ γ π π/π π=π/ππ₯ (π cosβ‘γπ₯+π sinβ‘γπ₯ γ γ ) =π π(cosβ‘π₯ )/ππ₯+π π(sinβ‘π₯ )/ππ₯ =π(γβsinγβ‘π₯ )+π(cosβ‘π₯ ) =βπ ππππ+π ππππ Now, (π^2 π¦)/(ππ₯^2 )=π/ππ₯ (ππ¦/ππ₯) (π ^π π)/(π π^π ) =π /π π (βπ ππππ+π ππππ) (π^2 π¦)/(ππ₯^2 ) =βπ π(sinβ‘π₯ )/ππ₯+π (π(cosβ‘π₯))/ππ₯ (π^2 π¦)/(ππ₯^2 ) =βπ(cosβ‘π₯ )+π(βsinβ‘π₯) (π^2 π¦)/(ππ₯^2 ) =βπ πππ β‘γπ₯βπ π ππβ‘π₯ γ (π ^π π)/(π π^π ) =β(π πππβ‘γπ+π πππβ‘π γ ) Putting y = π cosβ‘γπ₯+π sinβ‘γπ₯ γ γ (π ^π π)/(π π^π )=βπ (π ^π π)/(π π^π )+π=π β΄ Hence Verified