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

Find the solution of the following Second Order Differential Equations.
$$\displaystyle{4}{y}\text{}{4}{y}'+{y}={0}$$

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reinosodairyshm
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 $$\displaystyle{4}{y}\text{}{4}{y}+{y}={0}\ \text{is}\ {y}={A}{e}^{{-{\frac{{{x}}}{{{2}}}}}}+{B}{x}{e}^{{-{\frac{{{x}}}{{{2}}}}}}$$
###### Not exactly what you’re looking for?
SlabydouluS62
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.
karton

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