Stacie Worsley

Answered

2022-01-02

Non-Homogeneous Linear Differential Equation (The Method of Undetermined Coefficients).

Solve the equation$(D+1)y=\mathrm{sin}x$

Solve the equation

Answer & Explanation

Alex Sheppard

Expert

2022-01-03Added 36 answers

equation (1) D.w.r. to x

again D.w.r. to x

eq (1) and eq (3) put in eq (2)

equat the coefficient

P.I is

Alex Sheppard

Expert

2022-01-04Added 36 answers

Here A.E. is

${m}^{2}+1=0$ and its roots are $m=\pm i$

Hence$C.F.={C}_{1}\mathrm{cos}x+{C}_{2}\mathrm{sin}x$

Note that sin x is common in the C.F. and the R.H.S. of the given equation. ($\pm$ i is the root of the A.E.)

Therefore P.I. is y the form$yp=x(a\mathrm{cos}x+b\mathrm{sin}x)$ ...(1)

Since$\pm$ i is root of the A.E.

We have to find a and b such that${y}_{p}+{y}_{p}=\mathrm{sin}x$ ...(2)

From Eqn. (1)${y}_{p}^{\prime}=x(-a\mathrm{sin}x+b\mathrm{cos}x)+(a\mathrm{cos}x+b\mathrm{sin}x)$

${y}_{p}=x(-a\mathrm{cos}x-b\mathrm{sin}x)+(-a\mathrm{sin}x+b\mathrm{cos}x)-a\mathrm{sin}x+b\mathrm{cos}x$

$=x(-a\mathrm{cos}x-b\mathrm{sin}x)-2a\mathrm{sin}x+2b\mathrm{cos}x$

Then the given equation reduces to using the Eqn. (1)$x(-a\mathrm{cos}x-b\mathrm{sin}x)-2a\mathrm{sin}x+2b\mathrm{cos}x+x(a\mathrm{cos}x+b\mathrm{sin}x)=\mathrm{sin}x$

Equating the coefficients, we get

$i.e.,-2a\mathrm{sin}x+2b\mathrm{cos}x=\mathrm{sin}x$

$-2a=1,2b=0$

$a=\frac{-1}{2},b=0$

Thus P.I. is${y}_{p}=\frac{-1}{2}x\mathrm{cos}x$

$y=C.F.+P.I.$

$={C}_{1}\mathrm{cos}x+{C}_{2}\mathrm{sin}x-\frac{1}{2}x\mathrm{sin}x$ .

Hence

Note that sin x is common in the C.F. and the R.H.S. of the given equation. (

Therefore P.I. is y the form

Since

We have to find a and b such that

From Eqn. (1)

Then the given equation reduces to using the Eqn. (1)

Equating the coefficients, we get

Thus P.I. is

karton

Expert

2022-01-09Added 439 answers

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