Solve the initial value problem \[ \begin{array}{ll} \frac{\mathrm{d}}{\mathrm{d} x} y(x)+4 y(x)=\mathrm{e}^{(-4 x)}(-3+4 x) ; & y(0)=3 \\ y(x)= & \text { 圆 } \end{array} \]

The given differential equation is a first order linear non-homogeneous differential equation. We can solve it using an integrating factor.

The general form of a first order linear differential equation is: dy/dx + p(x)y = q(x)

Comparing this with the given differential equation, we have p(x) = 4 and q(x) = e^(-4x)(-3+4x)

The integrating factor is e^(∫p(x) dx) = e^(4x)

Multiplying every term of the equation by the integrating factor, we get:

e^(4x) dy/dx + 4e^(4x)y = e^(4x)e^(-4x)(-3+4x)

Simplifying, we have:

d/dx [e^(4x)y] = -3 + 4x

Integrating both sides with respect to x, we get:

e^(4x)y = -3x + 2x^2 + C

Solving for y, we get:

y = e^(-4x)(-3x + 2x^2 + C)

Now, using the initial condition y(0) = 3, we can solve for C:

3 = e^(0)(-30 + 20^2 + C) 3 = C

So, the solution to the differential equation is:

y = e^(-4x)(-3x + 2x^2 + 3)