Answer: The general solution is

y = A e^{-x} + B e^{-4x} - 1.40\sin(3x) - 0.20\cos(3x).

For large positive \(x\), the exponential terms tend to \(0\), so

\(y \approx -1.40\sin(3x) - 0.20\cos(3x)\).

This can be written in the form

\(y \approx R\sin(3x+\phi)\), where

\(R = 1.41\) and \(\phi = 3.28\) radians.

First solve the homogeneous equation

\(\dfrac{d^2y}{dx^2}+5\dfrac{dy}{dx}+4y=0\).

Its auxiliary equation is \(m^2+5m+4=0\), so

\((m+1)(m+4)=0\), hence \(m=-1,-4\).

Therefore the complementary function is

\(y_c=Ae^{-x}+Be^{-4x}\).

Now look for a particular integral of the form

\(y_p=P\sin 3x+Q\cos 3x\).

Then

\(y_p' = 3P\cos 3x - 3Q\sin 3x\),

\(y_p'' = -9P\sin 3x - 9Q\cos 3x\).

Substitute into the differential equation:

\((-9P\sin 3x-9Q\cos 3x)+5(3P\cos 3x-3Q\sin 3x)+4(P\sin 3x+Q\cos 3x).\)

Collecting coefficients gives

\((-5P-15Q)\sin 3x + (15P-5Q)\cos 3x.\)

This must equal \(10\sin 3x-20\cos 3x\), so

\(-5P-15Q=10,\qquad 15P-5Q=-20.\)

Dividing the first equation by \(-5\) gives \(P+3Q=-2\). Solving with the second equation:

\(P=-\frac75,\qquad Q=-\frac15.\)

So the general solution is

\(y=Ae^{-x}+Be^{-4x}-\frac75\sin 3x-\frac15\cos 3x.\)

For large positive \(x\), the exponential terms vanish, so

\(y\approx -\frac75\sin 3x-\frac15\cos 3x.\)

Write this as \(R\sin(3x+\phi)\). Since

\(R\sin(3x+\phi)=R(\sin 3x\cos\phi+\cos 3x\sin\phi),\)

we compare coefficients:

\(R\cos\phi=-\frac75,\qquad R\sin\phi=-\frac15.\)

Hence

\(R=\sqrt{\left(\frac75\right)^2+\left(\frac15\right)^2}=\frac{\sqrt{50}}5=\sqrt2\approx 1.41.\)

Also

\(\tan\phi=\frac{(-1/5)}{(-7/5)}=\frac17.\)

Since both \(\sin\phi\) and \(\cos\phi\) are negative, \(\phi\) lies in the third quadrant, so

\(\phi=\pi+\arctan\left(\frac17\right)\approx 3.28.\)

Thus, for large positive \(x\),

\(\boxed{y\approx 1.41\sin(3x+3.28)}\)

with \(R=1.41\) and \(\phi=3.28\) radians.