Not first, Not second, its third order

Calculus Level 5

If the solution of the differential equation

$\large{\frac{d^3 y}{dx^3}+2\frac{d^2 y}{dx^2}+\frac{dy}{dx}=e^{2x}+x^2+x}$ is

$\large{y=c_{1}+(c_{2}+c_{3}x)e^{Ax}+\frac{B}{C}e^{Dx}+\frac{E}{F}x^{G}+Hx^{I}-\frac{J}{K}x^{L}}$

where $c_{1},c_{2},c_{3}$ are constants.

$A,B,C,D,E,F,G,H,I,J,K,L$ are integers and all the fractions are co prime.

Find $A+B+C+D+E+F+G+H+I+J+K+L$

The answer is 39.

1 solution

Kartik Sharma
Aug 2, 2015

Ah! A really easy problem in disguise.

$y''' + 2y'' + y' = {e}^{2x} + {x}^{2} + x$

Now, using Laplace transform,

${s}^{3}\L (y) - {s}^{2}y(0) -s y'(0) - y''(0) + 2{s}^{2}\L (y) - 2s y(0) - 2y'(0) + s\L (y) -y(0) = \frac{1}{s-2} + \frac{2}{{s}^{3}} + \frac{1}{{s}^{2}}$

Or,

$\L (y) = \frac{\frac{1}{s-2} + \frac{2}{{s}^{3}} + \frac{1}{{s}^{2}} + ({s}^{2}+2s+1)y(0) + (s +2)y'(0) + y''(0)}{{s}^{3} + 2{s}^{2} + s}$

Now the inverse transform, I believe can be done by hand quite easily but for now, I just check out for it in W|A and give the result. (However, you can easily work it out for yourself. Just manipulate it in "good" partial fractions so that they reach the defined inverse transform results given in the Table. Check out there and you will get the result. Try it out!)

The result -

$\displaystyle y = {c}_{1} + ({c}_{2} + {c}_{3}x){e}^{-x} + \frac{{e}^{2x}}{18} + \frac{{x}^{3}}{3} + 4x - \frac{3{x}^{2}}{2}$

For the original problem, we could also integrate both sides to obtain a linear second order differential equation and solve for the complementary function and particular integrals.

John Frank - 3 years ago

$\L(u)$ denotes Laplace transform of $u$ . Why isn't it turning to LATEX?

Kartik Sharma - 5 years, 10 months ago

I think there's no such input in Latex. Try out $\mathcal{L} (u)$ .

Tapas Mazumdar - 3 years, 1 month ago

Not first, Not second, its third order

The answer is 39.

1 solution

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