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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] .
Find out more information about solution of differential equation here:
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