Calculus challenge.

Thank you Alan,  I shall study your solution,   

Here is my solution:

Solve:

$y''-2y'+y=(x^2+1)e^x$

Big Hint:

The solution will be of the from  

$y=(Ax^4+Bx^3+Cx^2)e^x \\~\\ \text{Where A,B and C are constants.}$

$y=(Ax^4+Bx^3+Cx^2)e^x\\~\\ y'=[Ax^4+(4A+B)x^3+(3B+C)x^2+2Cx]e^x\\~\\ y''=[Ax^4+(4A+4A+B)x^3+(12A+3B+3B+C)x^2+(6B+2C+2C)x+2C]e^x\\ y''=[Ax^4+(8A+B)x^3+(12A+6B+C)x^2+(6B+4C)x+2C]e^x\\~\\ y''-2y'+y\\ \begin{array}{rrrrrrr} =e^x[&Ax^4&+&(8A+B)x^3&+&(12A+6B+C)x^2&+&(6B+4C)x&+&2C&\\ &+(-2A)x^4&+&(-8A-2B)x^3&+&(-6B-2C)x^2&+&(-4C)x&\\ &+Ax^4&+&Bx^3&+&Cx^2&&&&&]\\~\\ =e^x[&0x^4&+&0x^3&+&12Ax^2&+&6Bx&+&2C&]\\~\\ \end{array}\\ so\\ y''-2y'+y=e^x[12Ax^2+6Bx+2C]\\~\\ $

$[12Ax^2+6Bx+2C]e^x=(x^2+1)e^x\\~\\ 12A=1 \qquad 6B=0 \qquad 2C=1\\ A=\frac{1 }{12}\qquad \;\; B=0 \qquad \;\; C=\frac{1}{2}\\~\\ so\;\;if\;\;\\ y''-2y'+y=(x^2+1)e^x\\ then\\ y=\left(\dfrac{x^4}{12}+\dfrac{x^2}{2}\right)e^x$

Alternative hint:  First multiply through by e^-x then make the substituion z = ye^-x

.

y = c₁e^x + c₂xe^x + x²e^x/2 + x⁴e^x/12

and now one for you Moderators....

Solve this simple ODE...

y^''+ e^-axy = 0   with y(0) = 0, y'(0) = 1

This has an exact analytical solution - so don't believe all the software that says it doesn't.  Good luck.

y''-2y'+y=(x^2+1)e^x

Using LaPLace Transform, you get this answer!!:

y(x) = c_2 e^x x + c_1 e^x + (e^x x^4)/12 + (e^x x^2)/2

Here's my worked solution:

.

Hi guest/s

Thanks for your participation here :)     

I shall look at your solution and your new question too :)

I might take a while to get too it but I will  :))       

D operators work well for the first equation.

$\displaystyle y'' - 2y' + y = (x^{2}+1)e^{x}$

so the equation becomes

$\displaystyle D^{2}y-2Dy + y =(x^{2}+1)e^{x}$,

$\displaystyle (D^{2}-2D+1)y=(x^{2}+1)e^{x}$,

$\displaystyle y = \frac{1}{(D-1)^{2}}e^{x}(x^{2}+1)=e^{x}\frac{1}{D^{2}}(x^{2}+1)=e^{x}\big(\frac{x^{4}}{12}+\frac{x^{2}}{2}+Ax+B\big)$,

where $\displaystyle A \text{ and }B$ are arbitrary constants.

Tiggsy