IDNLearn.com provides a platform for sharing and gaining valuable knowledge. Join our interactive Q&A platform to receive prompt and accurate responses from experienced professionals in various fields.
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:
https://brainly.com/question/2292185
#SPJ4
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.