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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 4 — Special Second-Ordered Differential Equations


Find the general solution of the differential equation

y+2y6x3=0y''+2y'-6x-3=0

Solution:

A second-order differential equation can be written in the form:

ay+by+cy=g(x)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+3y''+2y'=6x+3

ii. Let

y=P=dydxandy=dPdxdPdx+P2=6x+3\begin{align*} y' = P=\frac{dy}{dx} \\ \\ and \\ \\ y'' = \frac{dP}{dx} \\ \\ \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.

P(x)=2andQ(x)=6x+3dPdx+P2=6x+3\begin{align*} P(x) & =2 \\ and \\ Q(x) &=6x+3 \\ \\ \frac{dP}{dx}+P2\:& =\:6x+3 \end{align*}

Find the integrating factor

ɸ=eP(x)dxɸ=e  2dxɸ=e2x\begin{align*} ɸ & =e^{\int \:P\left(x\right)dx} \\ ɸ & =e^{\int \:\:2dx} \\ ɸ & =e^{2x} \end{align*}

Substituting the I.F. to the formula

Pɸ=ɸQ(x)dx+C1Pe2x=e2x(6x+3)dx+C1Pe2x=(e2x6x+3e2x)dx+C1\begin{align*} Pɸ & =\int \:ɸQ\left(x\right)dx+C_1 \\ Pe^{2x} & =\int \:e^{2x}(6x+3)dx+C_1 \\ Pe^{2x} & =\int \:\left(e^{2x}6x+3e^{2x}\right)dx+C_1 \end{align*}

Integrating the first term

e2x6xdx=6xe2xdx\begin{align*} \int \:e^{2x}6xdx= 6\cdot \int \:xe^{2x}dx \\ \end{align*}

Let u = 2x and du/2 = dx

32euudu\frac{3}{2}\int \:e^uudu

By IBP, Let v=u, dv=du and eudu, n=eu.

nvndvueu32  eudu=euueu=3e2xx32e2x\begin{align*} nv-\int ndv & \\ ue^u-\frac{3}{2}\int \:\:e^udu & = e^uu-e^u \\ & =3e^{2x}x-\frac{3}{2}e^{2x} \end{align*}

for the second term

3e2xdx\int \:3e^{2x}dx

Let u = 2x and du/2 = dx

32eudu=32eu=32e2x\begin{align*} \frac{3}{2}\int \:e^udu & =\frac{3}{2}e^u \\ & =\frac{3}{2}e^{2x} \end{align*}

Combining all the solved terms we get

Pe2x=3e2xx32e2x+32e2x+C1Pe2x=3e2xx+C1\begin{align*} Pe^{2x} & =3e^{2x}x-\frac{3}{2}e^{2x}+\frac{3}{2}e^{2x}+C_1 \\ Pe^{2x} & =3e^{2x}x+C_1 \end{align*}

Based on the equation that we derived it is now a separable differential equation, therefore,

[dydxe2x=3e2xx+C1]1e2xdydx=3x+C1e2xdy=(3x+C1e2x)dx+C2\begin{align*} &\left[\frac{dy}{dx}e^{2x}=3e^{2x}x+C_1\right]\frac{1}{e^{2x}} \\ \frac{dy}{dx} & =3x+\frac{C_1}{e^{2x}} \\ \int \:dy & =\int \:\left(3x+\frac{C_1}{e^{2x}}\right)dx+C_2 \end{align*}

GENERAL SOLUTION:

y=3x22C1e2x2+C2y=\frac{3x^2}{2}-\frac{C_1e^{-2x}}{2}+C_2

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