Answer:
[tex]y(x)=-\frac{6}{7}e^{-4x}+\frac{1}{7}e^{3x}[/tex] (See attached graph)
Step-by-step explanation:
Given Second-Order Homogeneous Differential Equation
[tex]y''+7y'+12y=0,y(0)=-1,y'(0)=0[/tex]
Use Auxiliary Equation
[tex]m^2+7m+12=0\\\\(m+4)(m+3)=0\\\\m=-4,\: m=3[/tex]
General Solution for Distinct Real Roots
[tex]y(x)=C_1e^{m_1x}+C_2e^{m_2x}\\\\y(x)=C_1e^{-4x}+C_2e^{3x}[/tex]
Take the derivative of y(x)
[tex]y'(x)=-4C_1e^{-4x}+3C_2e^{3x}[/tex]
Create a system of equations given initial conditions
[tex]y(x)=C_1e^{-4x}+C_2e^{3x}\\\\y(0)=C_1e^{-4(0)}+C_2e^{3(0)}=-1\\\\C_1+C_2=-1[/tex]
[tex]y'(x)=-4C_1e^{-4x}+3C_2e^{3x}\\\\y'(0)=-4C_1e^{-4(0)}+3C_2e^{3(0)}=0\\\\-4C_1+3C_2=0[/tex]
Solve the system of equations
[tex]\left \{ {{C_1+C_2=-1} \atop {-4C_1+3C_2=0}} \right.\\\\\left \{ {{4C_1+4C_2=-1} \atop {-4C_1+3C_2=0}} \right.\\\\7C_2=-1\\\\C_2=-\frac{1}{7}[/tex]
[tex]C_1+C_2=-1\\\\C_1-\frac{1}{7}=-1\\ \\C_1=-\frac{6}{7}[/tex]
Final Solution
[tex]y(x)=C_1e^{-4x}+C_2e^{3x}\\\\y(x)=-\frac{6}{7}e^{-4x}+\frac{1}{7}e^{3x}[/tex]
