x , we get = − Now, differentiating ( ) with respect to x , we have = e – x Substituting the values of , d y dy and y in the given differential equation, we get L.H.S. = e – x + (– e – x ) – . e – x = e – x – e – x = = R.H.S.. Therefore, the given function is a solution of the given differential equation.
Example Verify that the function y = a cos x + b sin x , where, a , b ∈ R is a solution of the differential equation Solution The given function is y = a cos x + b sin x Differentiating both sides of equation ( ) with respect to x , successively, we get dx = – a sin x + b cos x = – a cos x – b sin x Substituting the values of and y in the given differential equation, we get L.H.S. = (– a cos x – b sin x ) + ( a cos x + b sin x ) = = R.H.S. Therefore, the given function is a solution of the given differential equation. EXERCISE .
In each of the Exercises to verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: . y = e x + : y ″ – y ′ = . y = x + x + C : y ′ – x – = . y = cos x + C : y ′ + sin x = .
y = : y ′ = . y = A x : xy ′ = y ( x ≠ ) . y = x sin x : xy ′ = y + x ( x ≠ and x > y or x < – y ) . xy = log y + C : y ′ =