Question

If =2x+3x+d is the general solution for dy/dx +(4x+3)y =0 , determine the particular solution for the equation when y(2) =1.

Answer 1

Answer:

$y=e^{-2x^2-3x+14}$

Step-by-step explanation:

Determine the general solution

$\frac{dy}{dx}+(4x+3)y=0,\: y(2)=1\\ \\\frac{dy}{dx}+4xy+3y=0\\ \\\frac{dy}{dx}=-4xy-3y\\ \\dy=(-4xy-3y)dx\\\\\frac{1}{y}dy=(-4x-3)dx\\ \\\int {\frac{1}{y} } \, dy=\int{(-4x-3)} \, dx\\ \\ ln(|y|)=-2x^2-3x+C$

Solve for C given the initial condition y(2)=1

$ln(|1|)=-2(2)^2-3(2)+C\\\\0=-2(4)-6+C\\\\0=-8-6+C\\\\0=-14+C\\\\14=C$

Plug the value of C into the general solution equation

$ln(|y|)=-2x^2-3x+14\\\\y=e^{-2x^2-3x+14}$