verificame si este procedimiento matematico está correcto, si esta bien me lo dices, si no, corrigeme:

To solve the differential equation
y′′−8y′+25y=5x3e−x−7e−x
y′′−8y′+25y=5x3e−x−7e−x using the method of undetermined coefficients, we need to follow these steps:

    Find the general solution to the homogeneous equation: The homogeneous equation is
    y′′−8y′+25y=0
    y′′−8y′+25y=0. The characteristic equation is:
    r2−8r+25=0
    r2−8r+25=0 Solving this quadratic equation using the quadratic formula
    r=−b±b2−4ac2a
    r=2a−b±b2−4ac

​​, we get:
r=8±64−1002=8±−362=8±6i2=4±3i
r=28±64−100
​​=28±−36

    ​​=28±6i​=4±3i Therefore, the general solution to the homogeneous equation is:
    yh=e4x(C1cos⁡3x+C2sin⁡3x)
    yh​=e4x(C1​cos3x+C2​sin3x)

    Find a particular solution to the non-homogeneous equation: The right-hand side of the non-homogeneous equation is
    5x3e−x−7e−x
    5x3e−x−7e−x. We will use the method of undetermined coefficients to find a particular solution. Since
    e−x
    e−x is not a solution to the homogeneous equation, we assume a particular solution of the form:
    yp=(Ax3+Bx2+Cx+D)e−x
    yp​=(Ax3+Bx2+Cx+D)e−x We need to find the coefficients
    A
    A,
    B
    B,
    C
    C, and
    D
    D. First, we compute the first and second derivatives of
    yp
    yp​:
    yp′=(3Ax2+2Bx+C)e−x−(Ax3+Bx2+Cx+D)e−x=(−Ax3+(3A−B)x2+(2B−C)x+(C−D))e−x
    yp′​=(3Ax2+2Bx+C)e−x−(Ax3+Bx2+Cx+D)e−x=(−Ax3+(3A−B)x2+(2B−C)x+(C−D))e−x
    yp′′=(−3Ax2+(6A−2B)x+(2B−2C))e−x−(−Ax3+(3A−B)x2+(2B−C)x+(C−D))e−x
    yp′′​=(−3Ax2+(6A−2B)x+(2B−2C))e−x−(−Ax3+(3A−B)x2+(2B−C)x+(C−D))e−x
    yp′′=(Ax3+(−3A+3A−B)x2+(6A−2B−2B+C)x+(2B−2C−C+D))e−x
    yp′′​=(Ax3+(−3A+3A−B)x2+(6A−2B−2B+C)x+(2B−2C−C+D))e−x
    yp′′=(Ax3+(−B)x2+(6A−4B+C)x+(2B−3C+D))e−x
    yp′′​=(Ax3+(−B)x2+(6A−4B+C)x+(2B−3C+D))e−x Now, we substitute
    yp
    yp​,
    yp′
    yp′​, and
    yp′′
    yp′′​ into the original differential equation:
    (Ax3+(−B)x2+(6A−4B+C)x+(2B−3C+D))e−x−8((−Ax3+(3A−B)x2+(2B−C)x+(C−D))e−x)+25((Ax3+Bx2+Cx+D)e−x)=5x3e−x−7e−x
    (Ax3+(−B)x2+(6A−4B+C)x+(2B−3C+D))e−x−8((−Ax3+(3A−B)x2+(2B−C)x+(C−D))e−x)+25((Ax3+Bx2+Cx+D)e−x)=5x3e−x−7e−x Combining like terms, we get:
    (Ax3+(−B)x2+(6A−4B+C)x+(2B−3C+D)+(−8A)x3+(24A−8B)x2+(−16B+8C)x+(−8C+8D)+25Ax3+25Bx2+25Cx+25D)e−x=5x3e−x−7e−x
    (Ax3+(−B)x2+(6A−4B+C)x+(2B−3C+D)+(−8A)x3+(24A−8B)x2+(−16B+8C)x+(−8C+8D)+25Ax3+25Bx2+25Cx+25D)e−x=5x3e−x−7e−x
    ((A−8A+25A)x3+(−B+24A−8B+25B)x2+(6A−4B+C−16B+8C+25C)x+(2B−3C+D−8C+8D+25D))e−x=5x3e−x−7e−x
    ((A−8A+25A)x3+(−B+24A−8B+25B)x2+(6A−4B+C−16B+8C+25C)x+(2B−3C+D−8C+8D+25D))e−x=5x3e−x−7e−x
    (18Ax3+(24A+6B)x2+(6A−20B+34C)x+(2B−11C+34D))e−x=5x3e−x−7e−x
    (18Ax3+(24A+6B)x2+(6A−20B+34C)x+(2B−11C+34D))e−x=5x3e−x−7e−x By comparing coefficients, we get the system of equations:
    18A=5
    18A=5
    24A+6B=0
    24A+6B=0
    6A−20B+34C=0
    6A−20B+34C=0
    2B−11C+34D=−7
    2B−11C+34D=−7 Solving these equations step by step:
    A=518
    A=185​
    24(518)+6B=0  ⟹  12018+6B=0  ⟹  203+6B=0  ⟹  6B=−203  ⟹  B=−109
    24(185​)+6B=0⟹18120​+6B=0⟹320​+6B=0⟹6B=−320​⟹B=−910​
    6(518)−20(−109)+34C=0  ⟹  3018+2009+34C=0  ⟹  53+2009+34C=0  ⟹  159+2009+34C=0  ⟹  2159+34C=0  ⟹  34C=−2159  ⟹  C=−215306
    6(185​)−20(−910​)+34C=0⟹1830​+9200​+34C=0⟹35​+9200​+34C=0⟹915​+9200​+34C=0⟹9215​+34C=0⟹34C=−9215​⟹C=−306215​
    2(−109)−11(−215306)+34D=−7  ⟹  −209+2365306+34D=−7  ⟹  −680306+2365306+34D=−7  ⟹  1685306+34D=−7  ⟹  34D=−7−1685306  ⟹  34D=−2142306−1685306  ⟹  34D=−3827306  ⟹  D=−382710404
    2(−910​)−11(−306215​)+34D=−7⟹−920​+3062365​+34D=−7⟹−306680​+3062365​+34D=−7⟹3061685​+34D=−7⟹34D=−7−3061685​⟹34D=−3062142​−3061685​⟹34D=−3063827​⟹D=−104043827​ Therefore, the particular solution is:
    yp=(518x3−109x2−215306x−382710404)e−x
    yp​=(185​x3−910​x2−306215​x−104043827​)e−x

    Combine the general solution and the particular solution: The general solution to the non-homogeneous equation is:
    y=yh+yp=e4x(C1cos⁡3x+C2sin⁡3x)+(518x3−109x2−215306x−382710404)e−x
    y=yh​+yp​=e4x(C1​cos3x+C2​sin3x)+(185​x3−910​x2−306215​x−104043827​)e−x

The final answer is:
e4x(C1cos⁡3x+C2sin⁡3x)+(518x3−109x2−215306x−382710404)e−x
e4x(C1​cos3x+C2​sin3x)+(185​x3−910​x2−306215​x−104043827​)e−x