Find the solution of the following Second Order Differential Equations. 4y"+4y'+y=0

Bobbie Comstock 2021-12-31 Answered
Find the solution of the following Second Order Differential Equations.
\(\displaystyle{4}{y}\text{}{4}{y}'+{y}={0}\)

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Expert Answer

reinosodairyshm
Answered 2022-01-01 Author has 2532 answers
Calculation:
Convert \(\displaystyle{4}{y}\text{}{4}{y}'+{y}={0}\) into characteristics equation.
\(\displaystyle{4}{D}^{{{2}}}+{4}{D}+{1}={0}\)
\(\displaystyle{\left({2}{D}\right)}^{{{2}}}+{2}\cdot{2}{D}\cdot{1}+{1}^{{{2}}}={0}\)
\(\displaystyle{\left({2}{D}+{1}\right)}^{{{2}}}={0}\)
\(\displaystyle{D}=-{\frac{{{1}}}{{{2}}}}\)
Thus, the solution of differential equation \(\displaystyle{4}{y}\text{}{4}{y}+{y}={0}\ \text{is}\ {y}={A}{e}^{{-{\frac{{{x}}}{{{2}}}}}}+{B}{x}{e}^{{-{\frac{{{x}}}{{{2}}}}}}\)
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SlabydouluS62
Answered 2022-01-02 Author has 1786 answers
The auxiliary equation of this equation is:
\(\displaystyle{m}^{{{2}}}+{4}{m}+{m}={0}\)
On comparing this equation with the general quadratic equation,
\(\displaystyle{a}{x}^{{{2}}}+{b}{x}+{c}={0}\)
Then, \(\displaystyle{a}={1},{b}={4},{c}={1}\)
Now, use the quadratic form to find the value of m.
\(\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{b}^{{{2}}}-{4}{a}{c}}}}}{{{2}{a}}}}\)
\(\displaystyle{m}={\frac{{-{4}\pm\sqrt{{{4}^{{{2}}}-{4}{\left({1}\right)}{\left({4}\right)}}}}}{{{2}{\left({4}\right)}}}}\)
\(\displaystyle{m}={\frac{{-{4}\pm\sqrt{{{0}}}}}{{{8}}}}\)
\(\displaystyle{m}=-{\frac{{{4}}}{{{8}}}}-{0},-{\frac{{{4}}}{{{8}}}}+{0}\)
\(\displaystyle{m}=-{\frac{{{1}}}{{{2}}}},-{\frac{{{1}}}{{{2}}}}\)
Since values of m are equal, then the solution of the given differential equation is:
\(\displaystyle{y}={\left({c}_{{{1}}}+{c}_{{{2}}}{x}\right)}{e}^{{-{\frac{{{1}}}{{{2}}}}{x}}}\)
Where, \(\displaystyle{C}_{{{1}}},{C}_{{{2}}}\) are arbitrary constant.
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karton
Answered 2022-01-09 Author has 8454 answers

\(\begin{array}{} 4y''+4y'+y=0 \\4D^{2}+4D+1=0 \\(2D)^{2}+2 \cdot 2D \cdot 1+1^{2}=0 \\(2D+1)^{2}=0 \\D=-\frac{1}{2} \\4y''+4y+y=0\ \text{is}\ y=Ae^{-\frac{x}{2}}+Bxe^{-\frac{x}{2}} \end{array}\)

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