# Find a general solution to y''+4y'+3.75y=109\cos5x To solve this, the first thing I did was find the general solutionto the homogeneous equivalent, and got c_1e^{-5x/2}+c_2e^{3x/2} Then i used the form K\cos(wx)+M\sin(wx) and got -2.72\cos(5x)+2.56\sin(5x) as a solution of the nonhomogeneous ODE

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Find a general solution to $$\displaystyle{y}{''}+{4}{y}'+{3.75}{y}={109}{\cos{{5}}}{x}$$
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 $$\displaystyle{K}{\cos{{\left({w}{x}\right)}}}+{M}{\sin{{\left({w}{x}\right)}}}$$ and got $$\displaystyle-{2.72}{\cos{{\left({5}{x}\right)}}}+{2.56}{\sin{{\left({5}{x}\right)}}}$$ as a solution of the nonhomogeneous ODE

2021-03-27
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

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