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The process is the same as for the other problem. The only difference is that the particular solution is a third order polynomial instead of a second order polynomial.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewest$2y'' + 3y' + 5y = 7x^2 + 9x + 11$

The homogeneous equation is:

$2y'' + 3y' + 5y = 0$

The general solution to the homogeneous equation has the form:

$y = c \, e^{\alpha x}$

Plugging this back into the H equation yields a complex conjugate solution pair:

$2 \alpha^2 \, y + 3 \alpha y + 5y = 0 \\ 2 \alpha^2 + 3 \alpha + 5 = 0 \\ \alpha_1 = \frac{-3 - \sqrt{31} i}{4} \\ \alpha_2 = \frac{-3 + \sqrt{31} i}{4}$

The solution to the H equation is:

$y_H = c_1 e^{\alpha_1 x} + c_2 e^{\alpha_2 x}$

Now for the particular solution associated with the the polynomial, it is evident that this must be a second order polynomial:

$y_P = A x^2 + B x + C$

Plugging into the differential equation:

$2(2A) + 3(2A x + B) + 5(A x^2 + B x + C) = 7x^2 + 9x + 11 \\ 5A x^2 + (6A + 5B)x + 4A + 3B + 5C = 7x^2 + 9x + 11 \\ 5A = 7 \implies A = \frac{7}{5} \\ 6A + 5B = 9 \implies B = \frac{3}{25} \\ 4A + 3B + 5C = 11 \implies C = \frac{126}{125}$

The particular solution is:

$y_P = \frac{7}{5} x^2 + \frac{3}{25} x + \frac{126}{125}$

The total solution is:

$y = y_H + y_P = c_1 e^{\alpha_1 x} + c_2 e^{\alpha_2 x} + \frac{7}{5} x^2 + \frac{3}{25} x + \frac{126}{125} \\ \alpha_1 = \frac{-3 - \sqrt{31} i}{4} \\ \alpha_2 = \frac{-3 + \sqrt{31} i}{4}$

To solve for constants $c_1$ and $c_2$, solve initial-value equations for the function and its derivative

End of first partSuppose instead that the differential equation had been:

$2y'' + 3y' + 5y = 7x^3 + 11x^2 + 13x + 9$

The homogeneous solution is identical to the previous problem. The particular solution has the form:

$y_P = A x^3 + B x^2 + Cx + D$

Plugging into the differential equation:

$2(6 A x + 2B) + 3(3 A x^2 + 2 B x + C) + 5(A x^3 + B x^2 + Cx + D) = 7x^3 + 11x^2 + 13x + 9 \\ 5A = 7 \implies A = \frac{7}{5} \\ 9A + 5B = 11 \implies B = -\frac{8}{25} \\ 12A + 6B + 5C = 13 \implies C = -\frac{47}{125} \\ 4B + 3C + 5D = 9 \implies D = \frac{1426}{625}$

The total solution is therefore:

$y = y_H + y_P = c_1 e^{\alpha_1 x} + c_2 e^{\alpha_2 x} + \frac{7}{5} x^3 -\frac{8}{25} x^2 -\frac{47}{125} x + \frac{1426}{625} \\ \alpha_1 = \frac{-3 - \sqrt{31} i}{4} \\ \alpha_2 = \frac{-3 + \sqrt{31} i}{4}$

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Thank you and also how to solve this $2y’’+3y’+5y=7x^3+11x^2+13x+9$ ? Thanks

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The process is the same as for the other problem. The only difference is that the particular solution is a third order polynomial instead of a second order polynomial.

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Please I give me a solution for this

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How about this 2y’’+3y’+5y=7x^4 how to solve this? Thanks

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Since the general procedure is the same as for the other two, I'll leave this one to you

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Is this correct if 4th degree is $y'' = 8Ax^2+6Bx+2C$, $y'=4Ax^3+3Bx^2+2Cx+D$ and $y = Ax^4+Bx^3+Cx^2+Dx+E$? I hope its right. Thanks

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Correct, except that the first term in your $y''$ equation should have a coefficient of $12$ instead of $8$

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$y''=12Ax^2+6Bx+2C$ is this correct?

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for $x^5$ the $y_{p}$ is $y''=15Ax^3+12Bx^2+6Cx+2D$? This this correct Sir?

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Look again at the coefficient on your $x^3$ term

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I think its 18?

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