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Vikentia [17]
3 years ago
15

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

ven differential equation in the form L(y) = 0, where L is a linear operator with constant coefficients. If possible, factor L. (Use D for the differential operator.) Incorrect: Your answer is incorrect. y = ex + 1 Find a linear differential operator that annihilates the function g(x) = ex + 1. (Use D for the differential operator.) Solve the given differential equation by undetermined coefficients. y(x) =
Mathematics
1 answer:
kompoz [17]3 years ago
3 0

Answer:

The general solution is

y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

     + \frac{x^2}{2}

Step-by-step explanation:

Step :1:-

Given differential equation  y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

(D^4 -2D^3+D^2)y = e^x+1

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

                         m^4 -2m^3+m^2 = 0

                         m^2(m^2 -2m+1) = 0

(m^2 -2m+1) this is the expansion of (a-b)^2

                        m^2 =0 and (m-1)^2 =0

The roots are m=0,0 and m =1,1

complementary function is y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x

<u>Step 2</u>:-

The particular equation is    \frac{1}{f(D)} Q

P.I = \frac{1}{D^2(D-1)^2} e^x+1

P.I = \frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}

P.I = I_{1} +I_{2}

\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}

\frac{1}{D} means integration

\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx

applying in integration u v formula

\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx  } )\, dx

I_{1} = \frac{1}{D^2(D-1)^2} e^x

\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))

\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

I_{2}= \frac{1}{D^2(D-1)^2}e^{0x}

\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x

again integration  \frac{1}{D} x = \frac{x^2}{2!}

The general solution is y = y_{C} +y_{P}

         y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

      + \frac{x^2}{2!}

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<h3>Does the table represent a function?</h3>

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