Non-Homogeneous Linear Differential Equation (The Method of Undetermined Coefficients). Solve the equation (D+1)y=sin⁡x

Stacie Worsley

Stacie Worsley

Answered

2022-01-02

Non-Homogeneous Linear Differential Equation (The Method of Undetermined Coefficients).
Solve the equation (D+1)y=sinx

Answer & Explanation

Alex Sheppard

Alex Sheppard

Expert

2022-01-03Added 36 answers

(D2+1)y=sinx
m2+1=0
m2=1
m=±i
C.f=C1cosx+C2sinx
yt=x(acosx+bsinx) (1)
ytyt=sinx (2)
equation (1) D.w.r. to x
yt=x(asinx+bcosx)+(acosx+bsinx)
again D.w.r. to x
ytx(acosxbsinx)+(asinx+vcosx)±asinx+bcosx
=x(acosxbsinx)2asinx+2bcosx (3)
eq (1) and eq (3) put in eq (2)
x(acosxbsinx)2asinx+2bcosx+x(cosx+bsinx)=sinx
equat the coefficient
2asinx+2bcosx=sinx
2a=1
a=12
2b=0
b=0
P.I is yt=12xcosx
y=c1cosx+c2sinx12xsinx
Alex Sheppard

Alex Sheppard

Expert

2022-01-04Added 36 answers

Here A.E. is
m2+1=0 and its roots are m=±i
Hence C.F.=C1cosx+C2sinx
Note that sin x is common in the C.F. and the R.H.S. of the given equation. (± i is the root of the A.E.)
Therefore P.I. is y the form yp=x(acosx+bsinx) ...(1)
Since ± i is root of the A.E.
We have to find a and b such that yp+yp=sinx ...(2)
From Eqn. (1) yp=x(asinx+bcosx)+(acosx+bsinx)
yp=x(acosxbsinx)+(asinx+bcosx)asinx+bcosx
=x(acosxbsinx)2asinx+2bcosx
Then the given equation reduces to using the Eqn. (1) x(acosxbsinx)2asinx+2bcosx+x(acosx+bsinx)=sinx
Equating the coefficients, we get
i.e.,2asinx+2bcosx=sinx
2a=1,2b=0
a=12,b=0
Thus P.I. is yp=12xcosx
y=C.F.+P.I.
=C1cosx+C2sinx12xsinx.
karton

karton

Expert

2022-01-09Added 439 answers

Given: (D2+1)y=sinxP.I.=1D2+1sinx=11+1sinx=x12Dsinx=x2sinxdx=x2(cosx)P.I.=xcosx2

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