Question

How do you find all solutions of the differential equation $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-4y?$

Answer 1

To find the solution of the given second order differential equation, we have to begin with finding the characteristic equation of the given differential equation. Then we find the roots of the characteristic equations. The characteristic equation has two roots. Later, using these roots we find the solutions of the given differential equation.

Complete step-by-step solution:
We are given with a second order homogeneous differential equation with constant coefficients,
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-4y.......\left( 1 \right)$
Transpose $4y$ to the left side, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}+4y=0.......\left( 2 \right)$
This is a second order homogeneous differential equation with constant coefficients of the form $a\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+b\dfrac{dy}{dx}+cy=0.......\left( 3 \right)$ with $a=1,\,\,b=0$ and $c=4.$
The characteristic equation or auxiliary equation of the equation $\left( 3 \right)$ is given by,
$\Rightarrow a{{m}^{2}}+bm+c=0$
Therefore, the characteristic equation for the equation $\left( 2 \right)$ is,
$\Rightarrow {{m}^{2}}+4=0.......\left( 4 \right)$ since $a=1,\,\,b=0$ and $c=4.$
Now the root of the characteristic equation $\left( 4 \right)$ is found as
$\Rightarrow {{m}^{2}}=-4$ by transposing $4.$
$\Rightarrow m=\sqrt{-4}$ by taking square root.
Since $m=\sqrt{-4},$ the values of $m=\pm 2i$ $\left[ \sqrt{-4}=\sqrt{-1\times 4}=\sqrt{-1}\sqrt{4}=2\sqrt{-1}=\pm 2i,i=\sqrt{-1} \right]$
That is, the equation $\left( 4 \right)$ has two complex roots,
${{m}_{1}}=2i$ and ${{m}_{2}}=-2i$
We know that if the characteristic equation $a{{m}^{2}}+bm+c=0$ of a second order homogeneous differential equation of the form $a\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+b\dfrac{dy}{dx}+cy=0$ has complex roots ${{m}_{1}}=\alpha +i\beta$ and ${{m}_{2}}=\alpha -i\beta ,$ then the solution of the differential equation is of the form $y={{e}^{\alpha x}}\left[ A\cos \beta x+B\sin \beta x \right]$ where $A$ and $B$ are constants.
Thus, the solution of the equation $\left( 2 \right)$ is given by,
$\Rightarrow y={{e}^{0.x}}\left[ A\cos 2x+B\sin 2x \right]$ Since $\alpha =0$ and $\beta =2.$
Now this becomes,
$\Rightarrow y={{e}^{0}}\left[ A\cos 2x+B\sin 2x \right]$
From this we will get,
$\Rightarrow y=1\times \left[ A\cos 2x+B\sin 2x \right],$ where ${{e}^{0}}=1$
**Now we get the solution of the given second order differential equation as
$\Rightarrow y=A\cos 2x+B\sin 2x.$ **

Note: We have to know that (i) if the characteristic equation has two equal real roots $m={{m}_{1}}={{m}_{2}},$ then the solution of the second order differential equation has the form $y=\left( {{c}_{1}}+{{c}_{2}}x \right){{e}^{mx}},$ (ii) if the characteristic equation has two distinct real roots ${{m}_{1}},{{m}_{2}},$ then the solution of the differential equation has the form $y={{c}_{1}}{{e}^{{{m}_{1}}x}}+{{c}_{2}}{{e}^{{{m}_{2}}x}},$ (iii) if the characteristic equation has complex roots ${{m}_{1}}=\alpha +i\beta$ and ${{m}_{2}}=\alpha -i\beta ,$ then the solution of the differential equation is of the form $y={{e}^{\alpha x}}\left[ A\cos \beta x+B\sin \beta x \right],$ ${{c}_{1}},{{c}_{2}},A,B$ are constants.