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Solve the given differential equation by undetermined coefficients. y'' 4y' 4y = 2x 3

Sagot :

The solution of the differential equation is [tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex] .

According to the given question.

We have a differential equation

[tex]y^{"} + 4y^{'} + 4y = 2x^{3}[/tex]

The above differerntial equation acn be written as

[tex](D^{2} +4D+ 4)= 2x^{3}[/tex]

Now, the auxillary equation for the above differential equation is given by

[tex]m^{2} + 4m + 4 = 0[/tex]

[tex]\implies m^{2} + 2m + 2m + 4 = 0[/tex]

⇒ m (m + 2) + 2(m + 2) = 0

⇒ m(m + 2)(m + 2) = 0

Therefore,

[tex]C.F = (C_{1} + C_{2}x)e^{-2x}[/tex]

Now,

[tex]PI = \frac{1}{D^{2} +4D+4} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4(1+\frac{D^{2}+4D }{4} )} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4} [1+(\frac{D^{2}+4D }{4} )]^{-1} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4} [ 1 - (\frac{D^{2} +4D}{4} )+(\frac{D^{2} +4D}{4} )^{2} -(\frac{D^{2}+4D }{4}) ^{3} ...]2x^{3}[/tex]

[tex]\implies PI =\frac{1}{2} [ x^{3} -\frac{1}{4} (6x)-3x^{2} +3x^{2} -6][/tex]

[tex]\implies PI = \frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex]

Therefore, the solution of the differential equation will be

y = CI + PI

[tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex]

Hence, the solution of the differential equation is [tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex] .

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