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.

Short 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-by-step solution

Taking the given function as y

First of all, take the given function as,

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

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,

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

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.