The solution of the differential equation {d^{2} y}/{d x^{2}}-3 \frac{d y}{d x}+2 y=e^{5 x} is

Correct Answer is : y=c1ex+c2e2x+112e5x

y=c_{1} e^{x}+c_{2} e^{2 x}+\frac{1}{12} e^{5 x}

y=c_{1} e^{-x}+c_{2} e^{2 x}+\frac{1}{12} e^{5 x}

y=c_{1} e^{x}+c_{2} e^{-2 x}+\frac{1}{12} e^{5 x}

y-c_{1} e^{x}+c_{2} e^{2 x}+\frac{1}{12} e^{-5 x}

BPSC 53-55th Exam Prelims 17 Apr 2011 Solved Paper

Question : 78 of 150

Marks: +1, -0

The solution of the differential equation

{d^{2} y}/{d x^{2}}-3 \;{d y}/{d x}+2 y=e^{5 x}

Solution:

From the question, equation is

\;{d^{2} y}/{d x^{2}}-3 \;{d y}/{d x}+2 y=e^{5 x}

[ Let

\;{d y}/{d x}= {D}]

( {D}^{2}-3 {D}+2)=e^{5 x}

( {D}-1)( {D}-2) y=e^{4 x}

C.F. is

y=c_{1} e^{x}+c_{2} e^{2 x}

P.I.

=\;{1}/{( {D}-1)( {D}-2)} e^{5 x}

=[\;{1}/{( {D}-1)}-\;{1}/{( {D}-2)}] e^{5 x}

=\;{1}/{ {D}-2} e^{5 x}-\;{1}/{ {D}-1} e^{5 x}

=e^{2 x} ∫ e^{5 x} e^{-2 x} d x-e^{x} ∫ e^{5 x} e^{-x} d x

=e^{2 x} \;{e^{3 x}}/{3}-e^{x} \;{e^{4 x}}/{4}=\;{e^{5 x}}/{12}

The complete solution is

y= {C.F.}+ {P.I}

⇒ y=c_{1} e^{x}+c_{2} e^{2 x}+\;{1}/{12} {e}^{5 {x}}

Go to Question:

Prev Question

Next Question