Question:

Find a general solution to \(\displaystyle{y}{''}+{4}{y}'+{3.75}{y}={109}{\cos{{5}}}{x}\)

\(\displaystyle{\left({D}^{{2}}+{4}{D}+{3.72}\right)}{Y}={0}\)

\(\displaystyle{\left({D}+{2.5}\right)}{\left({D}+{1.5}\right)}{Y}={0}\)

\(\displaystyle{C}.{F}.={C}_{{1}}{E}^{{-{2.5}{x}}}+{C}_{{2}}{E}^{{{1.5}{X}}}\)

\(\displaystyle{C}.{F}.={C}_{{1}}{E}^{{-{2.5}{X}}}+{C}_{{2}}{E}^{{-{1.5}{X}}}\)

To solve this, the first thing I did was find the general solutionto the homogeneous equivalent, and got \(\displaystyle{c}_{{1}}{e}^{{-{5}\frac{{x}}{{2}}}}+{c}_{{2}}{e}^{{-{3}\frac{{x}}{{2}}}}\)

Then I used the form \(\displaystyle{K}{\cos{{\left({w}{x}\right)}}}+{M}{\sin{{\left({w}{x}\right)}}}\)

\(\displaystyle{\left[{\frac{{{1}}}{{{D}^{{2}}+{4}{D}+{3.75}}}}\right]}{109}{\cos{{\left({5}{X}\right)}}}={109}{\left[{\frac{{{1}}}{{-{5}^{{2}}+{4}{D}+{3.75}}}}\right]}{\cos{{\left({5}{X}\right)}}}\)

\(\displaystyle={109}{\left[{1}\cdot{\frac{{{4}{D}+{21.25}}}{{{4}{D}-{21.25}}}}{\left({4}{D}+{21.25}\right)}\right]}{\cos{{\left({5}{X}\right)}}}\)

\(\displaystyle={109}{\frac{{{4}{D}+{21.25}}}{{{16}{D}^{{2}}-{21.25}^{{2}}}}}{\cos{{\left({5}{X}\right)}}}\)

\(\displaystyle={109}{\left[{\frac{{{4}{D}+{21.25}}}{{{16}\cdot-{5}^{{2}}-{21.25}^{{2}}}}}\right]}{\cos{{\left({5}{X}\right)}}}\)

\(\displaystyle={109}{\frac{{-{4}\cdot{5}{\sin{{\left({5}{X}\right)}}}+{21.25}{\cos{{\left({5}{X}\right)}}}}}{{-{400}-{451.5625}}}}\)

\(\displaystyle={109}{\left[{20}{\sin{{\left({5}{X}\right)}}}-{21.25}{\cos{{\left({5}{X}\right)}}}\right]}{851.5625}\)

\(\displaystyle={2.56}{\sin{{\left({5}{X}\right)}}}-{2.72}{\cos{{\left({5}{X}\right)}}}\)

and got \(\displaystyle-{2.72}{\cos{{\left({5}{x}\right)}}}+{2.56}{\sin{{\left({5}{x}\right)}}}\) as a solution of the nonhomogeneous

Find a general solution to \(\displaystyle{y}{''}+{4}{y}'+{3.75}{y}={109}{\cos{{5}}}{x}\)

\(\displaystyle{\left({D}^{{2}}+{4}{D}+{3.72}\right)}{Y}={0}\)

\(\displaystyle{\left({D}+{2.5}\right)}{\left({D}+{1.5}\right)}{Y}={0}\)

\(\displaystyle{C}.{F}.={C}_{{1}}{E}^{{-{2.5}{x}}}+{C}_{{2}}{E}^{{{1.5}{X}}}\)

\(\displaystyle{C}.{F}.={C}_{{1}}{E}^{{-{2.5}{X}}}+{C}_{{2}}{E}^{{-{1.5}{X}}}\)

To solve this, the first thing I did was find the general solutionto the homogeneous equivalent, and got \(\displaystyle{c}_{{1}}{e}^{{-{5}\frac{{x}}{{2}}}}+{c}_{{2}}{e}^{{-{3}\frac{{x}}{{2}}}}\)

Then I used the form \(\displaystyle{K}{\cos{{\left({w}{x}\right)}}}+{M}{\sin{{\left({w}{x}\right)}}}\)

\(\displaystyle{\left[{\frac{{{1}}}{{{D}^{{2}}+{4}{D}+{3.75}}}}\right]}{109}{\cos{{\left({5}{X}\right)}}}={109}{\left[{\frac{{{1}}}{{-{5}^{{2}}+{4}{D}+{3.75}}}}\right]}{\cos{{\left({5}{X}\right)}}}\)

\(\displaystyle={109}{\left[{1}\cdot{\frac{{{4}{D}+{21.25}}}{{{4}{D}-{21.25}}}}{\left({4}{D}+{21.25}\right)}\right]}{\cos{{\left({5}{X}\right)}}}\)

\(\displaystyle={109}{\frac{{{4}{D}+{21.25}}}{{{16}{D}^{{2}}-{21.25}^{{2}}}}}{\cos{{\left({5}{X}\right)}}}\)

\(\displaystyle={109}{\left[{\frac{{{4}{D}+{21.25}}}{{{16}\cdot-{5}^{{2}}-{21.25}^{{2}}}}}\right]}{\cos{{\left({5}{X}\right)}}}\)

\(\displaystyle={109}{\frac{{-{4}\cdot{5}{\sin{{\left({5}{X}\right)}}}+{21.25}{\cos{{\left({5}{X}\right)}}}}}{{-{400}-{451.5625}}}}\)

\(\displaystyle={109}{\left[{20}{\sin{{\left({5}{X}\right)}}}-{21.25}{\cos{{\left({5}{X}\right)}}}\right]}{851.5625}\)

\(\displaystyle={2.56}{\sin{{\left({5}{X}\right)}}}-{2.72}{\cos{{\left({5}{X}\right)}}}\)

and got \(\displaystyle-{2.72}{\cos{{\left({5}{x}\right)}}}+{2.56}{\sin{{\left({5}{x}\right)}}}\) as a solution of the nonhomogeneous