Question

verify that each given function is a solution of the

differential equation.

6.′′\:+\:2y′\:−\:3y\:=\:0;\:y\left(t\right)\:=\:e^{−3t}

Answer 1

$y′′\:+\:2y′\:−\:3y\:=\:0;\:y\left(t\right)\:=\:e^{−3t}$

Replacing $y'$ by $D$ , $y''$ by $D^2$ , we get:

$D^2+2D-3=0\\ \Rightarrow D^2+3D-D-3=0\\ \Rightarrow D(D+3)-1(D+3)=0\\ \Rightarrow (D-1)(D+3)=0\\ \therefore D=1,-3\\$

$\frac{1}{y}\frac{dy}{dx}=1,-3\\ \Rightarrow \frac{dy}{y}=dx; \frac{dy}{y}=-3dx\\$

Integrating both sides, we get:

$ln(y)=x+c,-3x+c$

$\therefore y=ke^x,ke^{-3x}$

Now verifying the solution:

When $y=ke^x,$ we get:

$\frac{dy}{dx}=ke^x,\frac{d^2y}{dx^2}=ke^x$

Now, substituting, these in the given differential equation, we get:

$y''+2y'-3y=0\\ \Rightarrow ke^x+2ke^x-3ke^x=0 \\ \Rightarrow 0=0$

Hence Verified.

When $y=ke^{-3x}$ , we get:

$\frac{dy}{dx}=-3ke^{-3x},\frac{d^2y}{dx^2}=9ke^{-3x}$

Now, substituting, these in the given differential equation, we get:

$y''+2y'-3y=0\\ \Rightarrow 9ke^{-3x}-6ke^{-3x}-3ke^{-3x}=0 \\ \Rightarrow 0=0$

Hence Verified.

Hence, solutions are obtained and verified.