Question

Consider the first order differential equation a) Is this a linear system? In other words, show whether superposition and homogeneity are satisfied. b) What is the solution of this differential equation? Give the most general expression.

Answer 1

Answer:

Part a: The equation is a  linear first order differential equation as superposition and homogeneity is satisfied.

Part b: The solution of the differential equation is $y(t)=\frac{u(t)-e^{-kt} u(t)}{k}$

Step-by-step explanation:

Part a:

Let us suppose the differential equation is given as

$\frac{d}{dy} y(t) +k y(t)=u(t)$

For  a solution, it is given as

$\frac{d}{dy} y_1(t) +k y_1(t)=u_1(t)$

For another solution it is given as

$\frac{d}{dy} y_2(t) +k y_2(t)=u_2(t)$

Adding these two solution gives

$\frac{d}{dy} y_1(t) +k y_1(t)=u_1(t) \\+ \\\frac{d}{dy} y_2(t) +k y_2(t)=u_2(t)\\\\\frac{d}{dy} (y_1(t)+y_2(t)) +k (y_1(t)+y_2(t))=(u_1(t)+u_2(t))$

Which is also the solution of the equation and thus as superposition (additivity) and homogeneity is satisfied, so the equation is a  linear first order differential equation.

Part b

$\frac{d}{dy} y(t) +k y(t)=u(t)$

So taking Laplace on both sides

$s Y(s)+kY(s)=\frac{1}{s}$

Here

$u(t)=\frac{1}{s}$

Rearranging the equation gives

$Y(s)=\frac{1}{k(s+k)}\\Y(s)=\frac{1}{k}[\frac{1}{s}-\frac{1}{s+k}]$

Taking inverse Laplace gives

$y(t)=\frac{u(t)-e^{-kt} u(t)}{k}$

This is the solution of the differential equation.