# (A) To show that e2x and ex are two linearly independent solutions o.pdf

A

(A) To show that e2x and ex are two linearly independent solutions of the given diff. eqn. (1) To show that both solutions satisfy the given diff. eqn.i.e. y\"-3y\' +2y=0 Let us take y=e2x . y\' = 2e2x and y\" = 4e2x , Now plug these values in given diff eqn. 4e2x - 3*2e2x + 2e2x =0 Now similarly for y=ex ex-3ex+ex =0. (2) To show that both are linearly independent We calculate the value of their Wronskian. Which is If f(x)=e^x and g(x)=e^2x then, Wronskian (W) = Det ( f(x) f \'(x) ) ( g(x) g \'(x) ) If it is non zero then f(x) and g(x) are linearly independent for all x in the domain. In this case W = Det ( ex ex ) ( e2x 2e2x ) = e3x Which is non zero everywhere. So both solutions are linearly independent. (B) The auxiliary eqn for the given diff eqn is m2 -3m +2 =0 (m-1)(m-2)=0 m=1 and m=2 So Complimentary function = A em1 + B em2 = A ex + B e2x Where A and B are arbitrary constants. (C) Since The polynomial on Right hand side is of degree 2 , we know our particular integral will be of type: y = P x2 +Q x +R ( P, Q and R are arbitrary constt.) So y\' = 2P x + Q and y\" = 2P Plug these values into the diff eqn. and compare coeffecient of X^2 , X and Constant. We get P=2 and Q= 6 and R= 7 SO our particular integral is = 2x2 + 6x+7 (D) The General solution of given eqn = C.F. + P.I. = A ex + B e2x + 2x2 +6x+7 Where A and B are arbitrary constants. Solution (A) To show that e2x and ex are two linearly independent solutions of the given diff. eqn. (1) To show that both solutions satisfy the given diff. eqn.i.e. y\"-3y\' +2y=0 Let us take y=e2x . y\' = 2e2x and y\" = 4e2x , Now plug these values in given diff eqn. 4e2x - 3*2e2x + 2e2x =0 Now similarly for y=ex ex-3ex+ex =0. (2) To show that both are linearly independent We calculate the value of their Wronskian. Which is If f(x)=e^x and g(x)=e^2x then, Wronskian (W) = Det ( f(x) f \'(x) ) ( g(x) g \'(x) ) If it is non zero then f(x) and g(x) are linearly independent for all x in the domain. In this case W = Det ( ex ex ) ( e2x 2e2x ) = e3x Which is non zero everywhere. So both solutions are linearly independent. (B) The auxiliary eqn for the given diff eqn is m2 -3m +2 =0 (m-1)(m-2)=0 m=1 and m=2 So Complimentary function = A em1 + B em2 = A ex + B e2x Where A and B are arbitrary constants. (C) Since The polynomial on Right hand side is of degree 2 , we know our particular integral will be of type: y = P x2 +Q x +R ( P, Q and R are arbitrary constt.) So y\' = 2P x + Q and y\" = 2P Plug these values into the diff eqn. and compare coeffecient of X^2 , X and Constant. We get P=2 and Q= 6 and R= 7 SO our particular integral is = 2x2 + 6x+7 (D) The General solution of given eqn = C.F. + P.I. = A ex + B e2x + 2x2 +6x+7 Where A and B are arbitrary constants. .

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### (A) To show that e2x and ex are two linearly independent solutions o.pdf

• 1. (A) To show that e2x and ex are two linearly independent solutions of the given diff. eqn. (1) To show that both solutions satisfy the given diff. eqn.i.e. y"-3y' +2y=0 Let us take y=e2x . y' = 2e2x and y" = 4e2x , Now plug these values in given diff eqn. 4e2x - 3*2e2x + 2e2x =0 Now similarly for y=ex ex-3ex+ex =0. (2) To show that both are linearly independent We calculate the value of their Wronskian. Which is If f(x)=e^x and g(x)=e^2x then, Wronskian (W) = Det ( f(x) f '(x) ) ( g(x) g '(x) ) If it is non zero then f(x) and g(x) are linearly independent for all x in the domain. In this case W = Det ( ex ex ) ( e2x 2e2x ) = e3x Which is non zero everywhere. So both solutions are linearly independent. (B) The auxiliary eqn for the given diff eqn is m2 -3m +2 =0 (m-1)(m-2)=0 m=1 and m=2 So Complimentary function = A em1 + B em2 = A ex + B e2x Where A and B are arbitrary constants. (C) Since The polynomial on Right hand side is of degree 2 , we know our particular integral will be of type: y = P x2 +Q x +R ( P, Q and R are arbitrary constt.) So y' = 2P x + Q and y" = 2P Plug these values into the diff eqn. and compare coeffecient of X^2 , X and Constant. We get P=2 and Q= 6 and R= 7 SO our particular integral is = 2x2 + 6x+7 (D) The General solution of given eqn = C.F. + P.I. = A ex + B e2x + 2x2 +6x+7 Where A and B are arbitrary constants.
• 2. Solution (A) To show that e2x and ex are two linearly independent solutions of the given diff. eqn. (1) To show that both solutions satisfy the given diff. eqn.i.e. y"-3y' +2y=0 Let us take y=e2x . y' = 2e2x and y" = 4e2x , Now plug these values in given diff eqn. 4e2x - 3*2e2x + 2e2x =0 Now similarly for y=ex ex-3ex+ex =0. (2) To show that both are linearly independent We calculate the value of their Wronskian. Which is If f(x)=e^x and g(x)=e^2x then, Wronskian (W) = Det ( f(x) f '(x) ) ( g(x) g '(x) ) If it is non zero then f(x) and g(x) are linearly independent for all x in the domain. In this case W = Det ( ex ex ) ( e2x 2e2x ) = e3x Which is non zero everywhere. So both solutions are linearly independent. (B) The auxiliary eqn for the given diff eqn is m2 -3m +2 =0 (m-1)(m-2)=0 m=1 and m=2 So Complimentary function = A em1 + B em2 = A ex + B e2x Where A and B are arbitrary constants. (C) Since The polynomial on Right hand side is of degree 2 , we know our particular integral will be of type: y = P x2 +Q x +R ( P, Q and R are arbitrary constt.) So y' = 2P x + Q and y" = 2P Plug these values into the diff eqn. and compare coeffecient of X^2 , X and Constant. We get P=2 and Q= 6 and R= 7 SO our particular integral is = 2x2 + 6x+7 (D) The General solution of given eqn = C.F. + P.I. = A ex + B e2x + 2x2 +6x+7 Where A and B are arbitrary constants.
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