Find the general solution of the differential equation
y''+2y'-6x-3=0
Solution:
A second-order differential equation can be written in the form:
ay″ + by' + cy = g(x)
Therefore, the problem given is a second order linear equation.
Here are STEPS on how to get the general solution:
i. simplify the equation to a Second ODE Form
y''+2y'=6x+3
ii. Let
\begin{align*}
y' = P=\frac{dy}{dx}
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\\
and
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y'' = \frac{dP}{dx}
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\frac{dP}{dx}+P2\:=\:6x+3
\end{align*}iii. By recalling, we can see that the equation is in First-order linear differential equation form. Solving the simplified equation using FOLDE.
\begin{align*}
P(x) & =2
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and
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Q(x) &=6x+3
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\frac{dP}{dx}+P2\:& =\:6x+3
\end{align*}Find the integrating factor
\begin{align*}
ɸ & =e^{\int \:P\left(x\right)dx}
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ɸ & =e^{\int \:\:2dx}
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ɸ & =e^{2x}
\end{align*}Substituting the I.F. to the formula
\begin{align*}
Pɸ & =\int \:ɸQ\left(x\right)dx+C_1
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Pe^{2x} & =\int \:e^{2x}(6x+3)dx+C_1
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Pe^{2x} & =\int \:\left(e^{2x}6x+3e^{2x}\right)dx+C_1
\end{align*}Integrating the first term
\begin{align*}
\int \:e^{2x}6xdx=
6\cdot \int \:xe^{2x}dx
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\end{align*}Let u = 2x and du/2 = dx
\frac{3}{2}\int \:e^uuduBy IBP, Let v=u, dv=du and eudu, n=eu.
\begin{align*}
nv-\int ndv &
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ue^u-\frac{3}{2}\int \:\:e^udu & = e^uu-e^u
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& =3e^{2x}x-\frac{3}{2}e^{2x}
\end{align*}for the second term
\int \:3e^{2x}dxLet u = 2x and du/2 = dx
\begin{align*}
\frac{3}{2}\int \:e^udu & =\frac{3}{2}e^u
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& =\frac{3}{2}e^{2x}
\end{align*}Combining all the solved terms we get
\begin{align*}
Pe^{2x} & =3e^{2x}x-\frac{3}{2}e^{2x}+\frac{3}{2}e^{2x}+C_1
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Pe^{2x} & =3e^{2x}x+C_1
\end{align*}Based on the equation that we derived it is now a separable differential equation, therefore,
\begin{align*}
&\left[\frac{dy}{dx}e^{2x}=3e^{2x}x+C_1\right]\frac{1}{e^{2x}}
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\frac{dy}{dx} & =3x+\frac{C_1}{e^{2x}}
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\int \:dy & =\int \:\left(3x+\frac{C_1}{e^{2x}}\right)dx+C_2
\end{align*}GENERAL SOLUTION:
y=\frac{3x^2}{2}-\frac{C_1e^{-2x}}{2}+C_2