Solve the initial value problem.
2y” - 11y’ + 12y =0 ; where y(0) = 5 and y’(0) = 15
Given, "2y''-11y'+12y=0."
Auxiliary equation,
"2m^2-11m+12=0\\\\\n(m-4)(m-\\frac{3}{2})=0\\\\\nm=4,\\frac{3}{2}"
Therefore, general solution is
"y(x)=c_1e^{4x}+c_2e^{\\frac{3}{2}x}.\\\\\ny'=4c_1e^{4x}+\\frac{3}{2}c_2e^{\\frac{3}{2}x}"
Given, the initial condition "y(0)=5 \\space and \\space y'(0)=-15" .
"y(0)=c_1+c_2\\\\\n\\implies 5=c_1+c_2\\\\\nand\\\\\ny'(0)=4c_1+\\frac{3}{2}c_2 \\\\\n\\implies -15=\n4c_1+\\frac{3}{2}c_2"
Solving above two equations, we get
"c_1=-9, c_2=14"
Therefore, particular solution of the given IVP is "y(x)=-9e^{4x}+14e^{\\frac{3}{2}x}" .
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