Example 2 Verify that the function 𝑦=𝑒^(−3𝑥) is a solution of the differential equation (𝑑^2 𝑦)/(𝑑𝑥^2 )+𝑑𝑦/𝑑𝑥−6𝑦=0 𝑦=𝑒^(−3𝑥) 𝒅𝒚/𝒅𝒙=𝑑(𝑒^(−3𝑥) )/𝑑𝑥 𝑑𝑦/𝑑𝑥=〖−3 𝑒〗^(−3𝑥) (𝒅^𝟐 𝒚)/(𝒅𝒙^𝟐 )=𝑑/𝑑𝑥 (𝑑𝑦/𝑑𝑥) =𝑑(〖−3 𝑒〗^(−3𝑥) )/𝑑𝑥 =−3 𝑑(𝑒^(−3𝑥) )/𝑑𝑥 =−3 × (〖−3 𝑒〗^(−3𝑥) ) = 〖9 𝑒〗^(−3𝑥) Now, we have to verify (𝒅^𝟐 𝒚)/(𝒅𝒙^𝟐 )+𝒅𝒚/𝒅𝒙−𝟔𝒚=𝟎 Solving L.H.S (𝑑^2 𝑦)/(𝑑𝑥^2 )+𝑑𝑦/𝑑𝑥−6𝑦 Putting values = 〖9 𝑒〗^(−3𝑥)+(−3𝑒^(−3𝑥) )−6(𝑒^(−3𝑥) ) =〖9 𝑒〗^(−3𝑥)−3𝑒^(−3𝑥)−6𝑒^(−3𝑥) =〖9 𝑒〗^(−3𝑥)−9𝑒^(−3𝑥) =𝟎 = R.H.S ∴ Hence Verified

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.