Q14 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Show that $\varphi \left(\mathbf{x}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}} \mathbf{sin} \mathbf{x}\mathbf{+}{\mathbf{c}}_{\mathbf{2}} \mathbf{cos} \mathbf{x}\mathbf{,}$ is a solution to $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$ for any choice of the constants ${\mathbf{c}}_{\mathbf{1}}$ and ${\mathbf{c}}_2$. Thus, ${\mathbf{c}}_{\mathbf{1}} \mathbf{sin} \mathbf{x}\mathbf{+}{\mathbf{c}}_{\mathbf{2}} \mathbf{cos} \mathbf{x}\mathbf{,}$ is a two-parameter family of solutions to the differential equation.

Step-by-Step Solution

Answer

$\varphi \left(\mathrm{x}\right)={\mathrm{c}}_1 \sin \mathrm{x}+{\mathrm{c}}_2 \cos \mathrm{x},$ is a solution to $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\mathrm{y}=0$, for any choice of the constants ${\mathrm{c}}_1$ and ${\mathrm{c}}_2$. 

${\mathrm{c}}_1 \sin \mathrm{x}+{\mathrm{c}}_2 \cos \mathrm{x},$ is a two-parameter family of solutions to the differential equation.

Step 1: Taking the given function as y

First of all, take the given function as,

$\varphi \left(\mathrm{x}\right)=\mathrm{y}$

Step 2: Differentiating the given function concerning x

Differentiating concerning x,  

$\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{c}}_1 \cos \mathrm{x}-{\mathrm{c}}_2 \sin \mathrm{x}$

Again, differentiating concerning x,

 $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-{\mathrm{c}}_1 \sin \mathrm{x}-{\mathrm{c}}_2 \mathrm{cosx}$

Step 3: Simplification of the differential equation obtained in step 2

In step 2, we get $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-{\mathrm{c}}_1 \mathrm{sinx}-{\mathrm{c}}_2 \mathrm{cosx}$

$\begin{array}{l}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-\left({\mathrm{c}}_1 \sin \mathrm{x}+{\mathrm{c}}_2 \cos \mathrm{x}\right)\\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-\mathrm{y}\\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\mathrm{y}=0\end{array}$

Which is identical to the given differential equation.

Hence $\mathit{\varphi }\left(x\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{x}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{ }\mathbf{cos}\mathbf{ }\mathbf{x}\mathbf{,}$ is a solution to $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$, for any choice of the constants ${\mathbf{c}}_{\mathbf{1}}$ and ${\mathbf{c}}_{\mathbf{2}}$. Therefore, ${\mathbf{c}}_{\mathbf{1}} \mathbf{sin} \mathbf{x}\mathbf{+}{\mathbf{c}}_{\mathbf{2}} \mathbf{cos} \mathbf{x}$ is a two-parameter family of solutions to the given differential equation.