Answer:
[tex]y=e^{-2x^2-3x+14}[/tex]
Step-by-step explanation:
Determine the general solution
[tex]\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\\[/tex]
Solve for C given the initial condition y(2)=1
[tex]ln(|1|)=-2(2)^2-3(2)+C\\\\0=-2(4)-6+C\\\\0=-8-6+C\\\\0=-14+C\\\\14=C[/tex]
Plug the value of C into the general solution equation
[tex]ln(|y|)=-2x^2-3x+14\\\\y=e^{-2x^2-3x+14}[/tex]
