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

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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Alex Sheppard
$$\displaystyle{\left({D}^{{{2}}}+{1}\right)}{y}={\sin{{x}}}$$
$$\displaystyle{m}^{{{2}}}+{1}={0}$$
$$\displaystyle{m}^{{{2}}}=-{1}$$
$$\displaystyle{m}=\pm{i}$$
$$\displaystyle{C}.{f}={C}_{{{1}}}{\cos{{x}}}+{C}_{{{2}}}{\sin{{x}}}$$
$$\displaystyle{y}_{{{t}}}={x}{\left({a}{\cos{{x}}}+{b}{\sin{{x}}}\right)}$$ (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
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 $$\displaystyle{y}_{{{p}}}={\frac{{-{1}}}{{{2}}}}{x}{\cos{{x}}}$$
$$\displaystyle{y}={C}.{F}.+{P}.{I}.$$
$$\displaystyle={C}_{{{1}}}{\cos{{x}}}+{C}_{{{2}}}{\sin{{x}}}-{\frac{{{1}}}{{{2}}}}{x}{\sin{{x}}}$$.
karton

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