Solution. Given \(2\frac{d^2x}{dt^2}+5\frac{dx}{dt}-3x=0\) (1)

when \(t=0\) then \(x=4\) and \(\frac{dx}{dt}=9\)

Auxiliary equation

\(2m^2+5m-3=0\)

\(2m^2+6m-m-3=0\)

\((2m-1)(m+3)=0\)

\(m_1=-3\) and \(m_2=\frac{1}{2}\)

Auxiliary equation

Then solution \(x(t)=c_1e^{3t}+c_2e^{\frac{1}{2t}}\) (2)

\(\frac{dx}{dt}=3c_1e^{3t}+\frac{1}{2c}_2e^{\frac{t}{2}}\)

\(x(0)=4\)

\(c_1+c_2=4\) (3)

Also \(\frac{dx}{dt}=9\) when \(t=0\)

\(3c_1+\frac{c_2}{2}=9\)

\(6c_1+c_2=18\) (4)

Solving (3) and (4) we get

\(5c_1=14\Rightarrow c_1=\frac{14}{5}\)

And \(c_2=\frac{6}{5}\)

Then solution (2) will becomes.

\(x(t)=\frac{14}{5}e^{3t}+\frac{6}{5}e^{\frac{t}{2}}\)

when \(t=0\) then \(x=4\) and \(\frac{dx}{dt}=9\)

Auxiliary equation

\(2m^2+5m-3=0\)

\(2m^2+6m-m-3=0\)

\((2m-1)(m+3)=0\)

\(m_1=-3\) and \(m_2=\frac{1}{2}\)

Auxiliary equation

Then solution \(x(t)=c_1e^{3t}+c_2e^{\frac{1}{2t}}\) (2)

\(\frac{dx}{dt}=3c_1e^{3t}+\frac{1}{2c}_2e^{\frac{t}{2}}\)

\(x(0)=4\)

\(c_1+c_2=4\) (3)

Also \(\frac{dx}{dt}=9\) when \(t=0\)

\(3c_1+\frac{c_2}{2}=9\)

\(6c_1+c_2=18\) (4)

Solving (3) and (4) we get

\(5c_1=14\Rightarrow c_1=\frac{14}{5}\)

And \(c_2=\frac{6}{5}\)

Then solution (2) will becomes.

\(x(t)=\frac{14}{5}e^{3t}+\frac{6}{5}e^{\frac{t}{2}}\)