Question

Which of the following functions are solutions of the differential equation y" – 4y' + 3y = 0? CA. y(x) = xe-3x B. y(x) = et C. y(x) = x D. y(x) = 23x E. y(x) = 3x F. y(x) = e-* G. y(x) = 0

Answer 1

Answer:

$y(x)=c_1e^{x}+c_2e^{3x}$

$y(x)=e^x$

$y(x)=e^{3x}$

Option B is correct. $y=e^t$

Step-by-step explanation:

Given: y''-4y'+3y = 0

It is a linear differential equation.

First we make characteristic equation.

$m^2-4m+3=0$

Now factor the equation and we get

$(m-3)(m-1)$

$m=1,3$

Here, we have two distinct root.

$y(x)=c_1e^{m_1t}+c_2e^{m_2t}$

Solution of differential equation:

$y(x)=c_1e^{x}+c_2e^{3x}$

where, c₁ and c₂ are constants.

Hence, The solution of the given differential equation is $y(x)=c_1e^{x}+c_2e^{3x}$

$y(x)=e^x$

$y(x)=e^{3x}$