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
y′=P=dxdyandy′′=dxdPdxdP+P2=6x+3
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)andQ(x)dxdP+P2=2=6x+3=6x+3
Find the integrating factor
ɸɸɸ=e∫P(x)dx=e∫2dx=e2x
Substituting the I.F. to the formula
PɸPe2xPe2x=∫ɸQ(x)dx+C1=∫e2x(6x+3)dx+C1=∫(e2x6x+3e2x)dx+C1
Integrating the first term
∫e2x6xdx=6⋅∫xe2xdx
Let u = 2x and du/2 = dx
23∫euudu
By IBP, Let v=u, dv=du and eudu, n=eu.
nv−∫ndvueu−23∫eudu=euu−eu=3e2xx−23e2x
for the second term
∫3e2xdx
Let u = 2x and du/2 = dx
23∫eudu=23eu=23e2x
Combining all the solved terms we get
Pe2xPe2x=3e2xx−23e2x+23e2x+C1=3e2xx+C1
Based on the equation that we derived it is now a separable differential equation, therefore,
dxdy∫dy[dxdye2x=3e2xx+C1]e2x1=3x+e2xC1=∫(3x+e2xC1)dx+C2
GENERAL SOLUTION:
y=23x2−2C1e−2x+C2
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