# Solve the following IVP: \left\{\begin{matrix} 4y'' - 12y' + 9y = 16 e^{\frac{3}{2}t}\\ y(0) =...

## Question:

Solve the following IVP: {eq}\left\{\begin{matrix} 4y'' - 12y' + 9y = 16 e^{\frac{3}{2}t}\\ y(0) = 4; y'(0) = 10 \end{matrix}\right. {/eq}

## Linear Differential Equation with Constant Coefficients:

A differential equation is said to be LDE with constant coefficients if it can be written as

{eq}\left( {{D^n} + {a_1}{D^{n - 1}} + {a_2}{D^{^{n - 2}}}......... + {a_n}} \right)y = Q(x) {/eq}

where {eq}{a_{1,}}{a_{2,}}{a_{3,}}.....{a_n} {/eq} are constant coefficients and Q(x) is a function of x or constant.

If {eq}Q(x) \ne 0 {/eq} then above LDE is said to be non homogeneous LDE and

If {eq}Q(x) = 0 {/eq} then above LDE is said to be homogeneous LDE.

General Solution Of DE:

{eq}y(t) = {y_c} + {y_p} {/eq}

where {eq}{y_c} {/eq} is known as complimentary function(CF)

and {eq}{y_p} {/eq} is known as Particular Integral(PI).

Now consider auxiliary equation {eq}f(m) = 0. {/eq}

Let roots of auxiliary equation are {eq}m = {m_1},{m_2},.....{m_n}. {/eq}

then solution of DE is

{eq}y = {c_1}{e^{{m_1}x}} + {c_1}{e^{{m_2}x}} + {c_1}{e^{{m_3}x}}...... + {c_1}{e^{{m_n}x}} {/eq}

When roots are real and equal i.e.

{eq}m = {m_1} = {m_2} = ..... = {m_n}. {/eq}

Then solution of DE is

{eq}y = {e^{{m_1}x}}(1 + {c_1}x + {c_2}{x^2} + ....{c_n}{x^n}) {/eq}

When roots are equal and complex i.e

{eq}m = {m_1} \pm i{m_2} = {m_3} \pm i{m_4}. {/eq}

Then solution of DE is

{eq}y = {e^{{m_1}x}}\left[ {({c_1} + {c_2}x)\cos {m_2}x + ({c_3} + {c_4}x)\sin {m_2}x} \right] {/eq}

Particular integral:

{eq}{y_p} = \frac{1}{{f(D)}}Q(x) {/eq}

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Given differential equation can be written in the form of

{eq}\left( {4{D^2} - 12D + 9} \right)y = 16{e^{\frac{3}{2}t}} {/eq}

{eq}\left( {{D^2} -...