y''+y'+y=0, y(0)=1, y'(0)=0

Answer:[tex]y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )[/tex]Step-by-step explanation:A second order linear , homogeneous ordinary differential equation has form [tex]ay''+by'+cy=0[/tex].Given: [tex]y''+y'+y=0[/tex]Let [tex]y=e^{rt}[/tex] be it's solution.We get,[tex]\left ( r^2+r+1 \right )e^{rt}=0[/tex]Since [tex]e^{rt}\neq 0[/tex], [tex]r^2+r+1=0[/tex]{ we know that for equation [tex]ax^2+bx+c=0[/tex], roots are of form [tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex] }We get,[tex]y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}[/tex]For two complex roots [tex]r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta[/tex], the general solution is of form [tex]y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )[/tex]i.e [tex]y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )[/tex]Applying conditions y(0)=1 on [tex]e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )[/tex], [tex]c_1=1[/tex]So, equation becomes [tex]y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )[/tex]On differentiating with respect to t, we get[tex]y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )[/tex]Applying condition: y'(0)=0, we get [tex]0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}[/tex]Therefore,[tex]y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )[/tex]