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

Stacie Worsley 2022-01-02 Answered
Non-Homogeneous Linear Differential Equation (The Method of Undetermined Coefficients).
Solve the equation \(\displaystyle{\left({D}+{1}\right)}{y}={\sin{{x}}}\)

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Expert Answer

Alex Sheppard
Answered 2022-01-03 Author has 1510 answers
\(\displaystyle{\left({D}^{{{2}}}+{1}\right)}{y}={\sin{{x}}}\)
\(\displaystyle{m}^{{{2}}}+{1}={0}\)
\(\displaystyle{m}^{{{2}}}=-{1}\)
\(\displaystyle{m}=\pm{i}\)
\(\displaystyle{C}.{f}={C}_{{{1}}}{\cos{{x}}}+{C}_{{{2}}}{\sin{{x}}}\)
\(\displaystyle{y}_{{{t}}}={x}{\left({a}{\cos{{x}}}+{b}{\sin{{x}}}\right)}\) (1)
\(\displaystyle{y}_{{{t}}}\text{}{y}_{{{t}}}={\sin{{x}}}\) (2)
equation (1) D.w.r. to x
\(\displaystyle{y}_{{{t}}}'={x}{\left(-{a}{\sin{{x}}}+{b}{\cos{{x}}}\right)}+{\left({a}{\cos{{x}}}+{b}{\sin{{x}}}\right)}\)
again D.w.r. to x
\(\displaystyle{y}_{{{t}}}\text{}{x}{\left(-{a}{\cos{{x}}}-{b}{\sin{{x}}}\right)}+{\left(-{a}{\sin{{x}}}+{v}{\cos{{x}}}\right)}\pm{a}{\sin{{x}}}+{b}{\cos{{x}}}\)
\(\displaystyle={x}{\left(-{a}{\cos{{x}}}-{b}{\sin{{x}}}\right)}-{2}{a}{\sin{{x}}}+{2}{b}{\cos{{x}}}\) (3)
eq (1) and eq (3) put in eq (2)
\(\displaystyle{x}{\left(-{a}{\cos{{x}}}-{b}{\sin{{x}}}\right)}-{2}{a}{\sin{{x}}}+{2}{b}{\cos{{x}}}+{x}{\left({\cos{{x}}}+{b}{\sin{{x}}}\right)}={\sin{{x}}}\)
equat the coefficient
\(\displaystyle-{2}{a}{\sin{{x}}}+{2}{b}{\cos{{x}}}={\sin{{x}}}\)
\(\displaystyle-{2}{a}={1}\)
\(\displaystyle{a}=-{\frac{{{1}}}{{{2}}}}\)
\(\displaystyle{2}{b}={0}\)
\(\displaystyle{b}={0}\)
P.I is \(\displaystyle{y}_{{{t}}}={\frac{{{1}}}{{{2}}}}{x}{\cos{{x}}}\)
\(\displaystyle{y}={c}_{{{1}}}{\cos{{x}}}+{c}_{{{2}}}{\sin{{x}}}-{\frac{{{1}}}{{{2}}}}{x}{\sin{{x}}}\)
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Alex Sheppard
Answered 2022-01-04 Author has 1510 answers
Here A.E. is
\(\displaystyle{m}^{{{2}}}+{1}={0}\) and its roots are \(\displaystyle{m}=\pm{i}\)
Hence \(\displaystyle{C}.{F}.={C}_{{{1}}}{\cos{{x}}}+{C}_{{{2}}}{\sin{{x}}}\)
Note that sin x is common in the C.F. and the R.H.S. of the given equation. (\(\displaystyle\pm\) i is the root of the A.E.)
Therefore P.I. is y the form \(\displaystyle{y}{p}={x}{\left({a}{\cos{{x}}}+{b}{\sin{{x}}}\right)}\) ...(1)
Since \(\displaystyle\pm\) i is root of the A.E.
We have to find a and b such that \(\displaystyle{y}_{{{p}}}\text{}+{y}_{{{p}}}={\sin{{x}}}\) ...(2)
From Eqn. (1) \(\displaystyle{y}_{{{p}}}'={x}{\left(-{a}{\sin{{x}}}+{b}{\cos{{x}}}\right)}+{\left({a}{\cos{{x}}}+{b}{\sin{{x}}}\right)}\)
\(\displaystyle{y}_{{{p}}}\text{}={x}{\left(-{a}{\cos{{x}}}-{b}{\sin{{x}}}\right)}+{\left(-{a}{\sin{{x}}}+{b}{\cos{{x}}}\right)}-{a}{\sin{{x}}}+{b}{\cos{{x}}}\)
\(\displaystyle={x}{\left(-{a}{\cos{{x}}}-{b}{\sin{{x}}}\right)}-{2}{a}{\sin{{x}}}+{2}{b}{\cos{{x}}}\)
Then the given equation reduces to using the Eqn. (1) \(\displaystyle{x}{\left(-{a}{\cos{{x}}}-{b}{\sin{{x}}}\right)}-{2}{a}{\sin{{x}}}+{2}{b}{\cos{{x}}}+{x}{\left({a}{\cos{{x}}}+{b}{\sin{{x}}}\right)}={\sin{{x}}}\)
Equating the coefficients, we get
\(\displaystyle{i}.{e}.,-{2}{a}{\sin{{x}}}+{2}{b}{\cos{{x}}}={\sin{{x}}}\)
\(\displaystyle-{2}{a}={1},{2}{b}={0}\)
\(\displaystyle{a}={\frac{{-{1}}}{{{2}}}},{b}={0}\)
Thus P.I. is \(\displaystyle{y}_{{{p}}}={\frac{{-{1}}}{{{2}}}}{x}{\cos{{x}}}\)
\(\displaystyle{y}={C}.{F}.+{P}.{I}.\)
\(\displaystyle={C}_{{{1}}}{\cos{{x}}}+{C}_{{{2}}}{\sin{{x}}}-{\frac{{{1}}}{{{2}}}}{x}{\sin{{x}}}\).
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karton
Answered 2022-01-09 Author has 8454 answers

\(\begin{array}{}Given:\ (D^{2}+1)y=\sin x \\P.I.=\frac{1}{D^{2}+1} \sin x \\=\frac{1}{-1+1} \sin x \\=x \frac{1}{2D} \sin x \\=\frac{x}{2} \int \sin x dx \\=\frac{x}{2} (-\cos x) \\P.I.=\frac{-x \cos x}{2} \end{array}\)

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