Question

Find a particular solution for${\mathit{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{\lambda }{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathit{A}\mathbf{,}$where A is a constant force.

Short answer

$x_p=\frac{A}{{\omega }^2}\mid$

Step-by-step solution

Step 1:Definition of Non-linear and Linear Spring

NONLINEAR SPRINGS The mathematical model has the form

$m\frac{d^{2}x}{d{t}^2}+F\left(x\right)=0$

where$F\left(x\right)=kx$. Becausedenotes the displacement of the mass from its equilibrium position,$F\left(x\right)=kx$is Hooke's law-that is, the force exerted by the spring that tends to restore the mass to the equilibrium position. A spring acting under a linear restoring force$F\left(x\right)=kx$is naturally referred to as a linear spring.

A spring whose mathematical model incorporates a nonlinear restorative force, such as

$m\frac{d^{2}x}{d{t}^2}+k{x}^3=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}m\frac{d^{2}x}{d{t}^2}+kx+k_1{x}^3=0$

is called a nonlinear spring.

Find the Particular solution

Since, the right hand side of the given equation is a constant, let us consider the constant function $x=c$as the particular solution of the given equation. So, we have $x^{\text{'}}=x^{\text{'}\text{'}}=0$. Substituting these in the equation gives

$x^{\text{'}\text{'}}+2\lambda x^{\text{'}}+{\omega }^{2}x=A$

$0+0+{\omega }^{2}c=A$

$c=\frac{A}{{\omega }^2}$

Hence, by inspection, the particular solution of$x^{\text{'}\text{'}}+2\lambda x^{\text{'}}+{\omega }^{2}x=A$is

$x_p=\frac{A}{{\omega }^2}\mid$