Question

Solve the given DE dx/dt=3x+2y+1 dy/dt=-2x-y+1

Answer 1

We can solve for $x(t)$ first by rewriting the system of first-order ODEs as a single second-order ODE in $x(t)$:

Taking the derivative of the first ODE gives

$\dfrac{\mathrm dx}{\mathrm dt}=3x+2y+1\implies\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}+2\dfrac{\mathrm dy}{\mathrm dt}$

while solving for $2y$ gives

$\dfrac{\mathrm dx}{\mathrm dt}=3x+2y+1\implies2y=\dfrac{\mathrm dx}{\mathrm dt}-3x-1$

Then

$\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}+2(-2x-y+1)$

$\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}-4x-2y+2$

$\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}-4x-\left(\dfrac{\mathrm dx}{\mathrm dt}-3x-1\right)+2$

$\implies\dfrac{\mathrm d^2x}{\mathrm dt^2}-2\dfrac{\mathrm dx}{\mathrm dt}+x=3$

which is linear with constant coefficients, so it's trivial to solve; the corresponding homogeneous ODE

$x''-2x'+x=0$

has characteristic equation

$r^2-2r+1=(r-1)^2=0$

with root $r=1$ (multiplicity 2), so the characteristic solution is

$x_c=C_1e^t+C_2te^t$

For the non-homogeneous ODE, assume a particular solution of the form

$x_p=a\implies{x_p}'={x_p}''=0$

Substituting these into the ODE gives

$0-2\cdot0+a=3\implies a=3$

Then the general solution for $x(t)$ is

$\boxed{x(t)=C_1e^t+C_2te^t+3}$

From here, we find

$\dfrac{\mathrm dx}{\mathrm dt}=C_1e^t+C_2(t+1)e^t$

so that

$2y=(C_1e^t+C_2(t+1)e^t)-3(C_1e^t+C_2te^t+3)-1$

$\implies\boxed{y(t)=\left(\dfrac{C_2}2-C_1\right)e^t-C_2te^t-5}$