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What is the solution for 2y+3y+5y=7x2+9x+112y''+3y'+5y=7x^2+9x+11 thanks.

Note by Ce Die
1 month, 1 week ago

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2y+3y+5y=7x2+9x+112y'' + 3y' + 5y = 7x^2 + 9x + 11

The homogeneous equation is:

2y+3y+5y=02y'' + 3y' + 5y = 0

The general solution to the homogeneous equation has the form:

y=ceαxy = c \, e^{\alpha x}

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

2α2y+3αy+5y=02α2+3α+5=0α1=331i4α2=3+31i42 \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:

yH=c1eα1x+c2eα2xy_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:

yP=Ax2+Bx+Cy_P = A x^2 + B x + C

Plugging into the differential equation:

2(2A)+3(2Ax+B)+5(Ax2+Bx+C)=7x2+9x+115Ax2+(6A+5B)x+4A+3B+5C=7x2+9x+115A=7    A=756A+5B=9    B=3254A+3B+5C=11    C=1261252(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:

yP=75x2+325x+126125y_P = \frac{7}{5} x^2 + \frac{3}{25} x + \frac{126}{125}

The total solution is:

y=yH+yP=c1eα1x+c2eα2x+75x2+325x+126125α1=331i4α2=3+31i4y = 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 c1 c_1 and c2 c_2 , solve initial-value equations for the function and its derivative

End of first part

Suppose instead that the differential equation had been:

2y+3y+5y=7x3+11x2+13x+92y'' + 3y' + 5y = 7x^3 + 11x^2 + 13x + 9

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

yP=Ax3+Bx2+Cx+Dy_P = A x^3 + B x^2 + Cx + D

Plugging into the differential equation:

2(6Ax+2B)+3(3Ax2+2Bx+C)+5(Ax3+Bx2+Cx+D)=7x3+11x2+13x+95A=7    A=759A+5B=11    B=82512A+6B+5C=13    C=471254B+3C+5D=9    D=14266252(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=yH+yP=c1eα1x+c2eα2x+75x3825x247125x+1426625α1=331i4α2=3+31i4y = 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}

Steven Chase - 1 month ago

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Thank you and also how to solve this 2y+3y+5y=7x3+11x2+13x+92y’’+3y’+5y=7x^3+11x^2+13x+9 ? Thanks

Ce Die - 1 month ago

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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.

Steven Chase - 1 month ago

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Please I give me a solution for this

Ce Die - 4 weeks, 1 day ago

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@Ce Die Ok, I will post it a bit later

Steven Chase - 4 weeks, 1 day ago

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@Steven Chase thank you i hope to see it

Ce Die - 4 weeks ago

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@Ce Die I have added it

Steven Chase - 4 weeks ago

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@Steven Chase thank you so much you enlight me! :)

Ce Die - 3 weeks, 3 days ago

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@Ce Die You're welcome

Steven Chase - 3 weeks, 3 days ago

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

Ce Die - 1 week, 5 days ago

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

Steven Chase - 1 week, 5 days ago

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Is this correct if 4th degree is y=8Ax2+6Bx+2Cy'' = 8Ax^2+6Bx+2C, y=4Ax3+3Bx2+2Cx+Dy'=4Ax^3+3Bx^2+2Cx+D and y=Ax4+Bx3+Cx2+Dx+Ey = Ax^4+Bx^3+Cx^2+Dx+E? I hope its right. Thanks

Ce Die - 1 week, 3 days ago

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Correct, except that the first term in your yy'' equation should have a coefficient of 1212 instead of 88

Steven Chase - 1 week, 3 days ago

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y=12Ax2+6Bx+2Cy''=12Ax^2+6Bx+2C is this correct?

Ce Die - 1 week, 3 days ago

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@Ce Die Yes, it is

Steven Chase - 1 week, 3 days ago

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for x5x^5 the ypy_{p} is y=15Ax3+12Bx2+6Cx+2D y''=15Ax^3+12Bx^2+6Cx+2D? This this correct Sir?

Ce Die - 1 week, 3 days ago

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Look again at the coefficient on your x3x^3 term

Steven Chase - 1 week, 2 days ago

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I think its 18?

Ce Die - 1 week, 2 days ago

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