Solve the differential equation:(D2 + 4D + 1)y = sin x .

1.	Solve the differential equation:(D2 + 4D + 1)y = sin x .
Answer» (D² + 4D + 1)y = sin x It's auxiliary equation is m² + 4m + 1 = 0 = m² + 4m + 4 - 4 + 1 = 0 = (m + 2)²= 3 = m + 2 = ± √3 = m = -2 ± √3 C.F = C₁e^{-(2 + √3)x} + C₂e^{-(2 - √3)x} P.I = \(\cfrac1{D^2+4D+1}\) sin x = \(\cfrac1{-1^2+4D+1}\)sin x = \(\cfrac1{4D}\) sin x = \(\cfrac1{4}\) \(\int\) sin x (\(\because\) \(\cfrac1{D}\) f(x) = \(\int\) f(x) ) = - \(\cfrac1{4}\) cos x \(\therefore\) complete solution of given differential equation is y = C.F + P.I = C₁e^{-(2 + √3)x} + C₂e^{-(2 - √3)x} ^{\(\cfrac{-cos\,x}4\)}

Answer»

(D² + 4D + 1)y = sin x

It's auxiliary equation is

m² + 4m + 1 = 0

= m² + 4m + 4 - 4 + 1 = 0

= (m + 2)²= 3

= m + 2 = ± √3

= m = -2 ± √3

C.F = C₁e^{-(2 + √3)x} + C₂e^{-(2 - √3)x}

P.I = \(\cfrac1{D^2+4D+1}\) sin x

= \(\cfrac1{-1^2+4D+1}\)sin x

= \(\cfrac1{4D}\) sin x

= \(\cfrac1{4}\) \(\int\) sin x (\(\because\) \(\cfrac1{D}\) f(x) = \(\int\) f(x) )

= - \(\cfrac1{4}\) cos x

\(\therefore\) complete solution of given differential equation is

y = C.F + P.I

= C₁e^{-(2 + √3)x} + C₂e^{-(2 - √3)x}

^{\(\cfrac{-cos\,x}4\)}

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