Example 2 - Chapter 9 Class 12 Differential Equations
Last updated at Dec. 16, 2024 by Teachoo
Gen and Particular Solution
Gen and Particular Solution
Last updated at Dec. 16, 2024 by Teachoo
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