Find the general solution for the nonhomogeneous differential equation:

{eq}y" + 3y' - 10y = 6e^{2x} {/eq}

Given:

• Consider the given non-homogeneous differential equation {eq}y'' + 3y' - 10y = 6{e^{2x}}{/eq} .

Let us now solve the above differential equation.

Let {eq}D = \dfrac{d}{{dx}},{D^2} = \dfrac{{{d^2}}}{{dx}}{/eq} . by plugging this in the given equation, we get,

{eq}\left( {{D^2} + 3D - 10} \right)y = 6{e^{2x}}{/eq}

First, we shall find the complementary function as follows,

Let {eq}D = m{/eq}, then to find the complementary function, consider

{eq}{m^2} + 3m - 10 = 0{/eq}

By factorizing we get,

{eq}m = - 5,2{/eq}

Therefore the complementary function is

{eq}{y_1} = A{e^{ - 5x}} + B{e^{2x}}{/eq}

Now let us find the particular integral for {eq}6{e^{2x}}{/eq} .

As {eq}f\left( a \right) = 0{/eq} we find that the particular integral is of the form {eq}Cx{e^{2x}}{/eq} , {eq}C{/eq} is a constant.

Let us now find {eq}C{/eq},

Consider,

{eq}{y_2} = Cx{e^{2x}}{/eq}

Differentiating with respect to {eq}x{/eq} , we get

{eq}{y_2}' = 2Cx{e^{2x}} + C{e^{2x}}{/eq}

Again differentiating,

{eq}{y_2}'' = 4Cx{e^{2x}} + 4C{e^{2x}}{/eq}

Substituting these values in the given equation we get,

{eq}\begin{align*} \left( {4Cx{e^{2x}} + 4C{e^{2x}}} \right) + 3\left( {2Cx{e^{2x}} + C{e^{2x}}} \right) - 10Cx{e^{2x}} &= 6{e^{2x}}\\ 7C{e^{2x}} &= 6{e^{2x}}\\ C &= \dfrac{6}{7} \end{align*}{/eq}

Thus,

{eq}{y_2} = \dfrac{6}{7}x{e^{2x}}{/eq}

Therefore the solution is {eq}y = {y_1} + {y_2}{/eq}

Hence, {eq}y = A{e^{ - 5x}} + {e^{2x}}\left( {B + \dfrac{6}{7}} \right){/eq} .