Test: Linear Differential Equations

dy/dx + 3y = 2e³x
p = 3,    q = 2e³x
∫p.dx = 3x
Integrating factor (I.F) = e³x
y(I.F) = ∫Q(I.F) dx
ye³x = ∫2e³x e³x dx
ye³x = ∫2e⁶x dx
ye³x = 2∫e⁶x dx
ye³x = 2/6[e⁶x] + c
ye³x = ⅓[e⁶x] + c
Dividing by e³x, we get
 y = ⅓[e³x] + ce^-(3x)

xlog x dy/dx + y = 2logx
⇒ dy/dx + y/(xlogx) = 2/x...(1)
Put P = 1/x logx
⇒∫PdP = ∫1/x logx dx 
= log(logx)
∴ I.F.= e^∫PdP = e^log(logx)
 = logx

xdy = (2y + 2x⁴ + x²)dx
→ dy/dx − (2x)y = 2x³ + x
This differential is of the form y′+P(x)y=Q(x) which is the general first order linear differential equation, where P(x) and Q(x) are continuous function defined on an interval.
The general solution for this is y∙I.F = ∫I.F × Q(x)dx
Where I.F = e∫P(x)dx is the integrating factor of the differential equation.
I.F = e∫P(x)dx
= e^∫−2/xdx
= e(−2∙lnx)
= e^ln(x⁻²)
= x⁻²
Thus y(1/x²) = ∫1/x²(2x³ + x)dx
=∫(2x + 1/x)dx
= x² + lnx + C
⟹ y = x⁴ + x²lnx + c

 dy/dx+2ytanx=sinx
This is in the form of dy/dx + py = θ
where p=2tanx,θ=sinx
∴ finding If e^∫pdx = e^∫2tanxdx 
=e^{(2log secx)}
=e^{(log sec2x)}
=sec²x

dy/dx + y/x = x²
differential equation is in the form : dy/dx + Py = Q
P = 1/x      Q = x²
I.F = e∫P(x)dx
I.F = e∫1/x dx
I.F = e[log x] 
I.F = x
y I.F = ∫(Q * I.F) dx + c
yx = ∫x² * x * dx + c
yx = ∫x³ dx + c
xy = [x⁴]/4 + c 

The given differential equation may be written as
dy/dx + (2x/(x²+1)y = √x2+4/(x²+1)  ... (i)
This is of the form dy/dx + Py = Q,
 where P=2x/(x²+1) and Q=(√x²+4)/(x²+1)
Thus, the given differential equation is linear.
IF=e^(∫Pdx)
= e(∫2x(x²+1)dx)
= e(log(x²+1) = (x²+1)
So, the required solution is given by
y × IF = ∫{Q×IF}dx + C,
i.e., y(x²+1)=∫(√x²+4)/(x²+1)×(x²+1)dx
⇒y(x²+1)=∫(√x²+4)dx
=1/2x (√x²+4) +1/2 × (2)² × log|x+(√x²+4)| + C
=1/2x (√x²+4) + 2log|x+(√x²+4) + C.
Hence, y(x²+1) = 1/2x(√x²+4) + 2log|x+√x²+4| + C is the required solution.

dy/dx + y/x = x
Differential eq is in the form of : dy/dx + Py = Q
P = 1/x     Q = x
I.F. = e^∫Pdx
= e^∫(1/x)dx
= e^{(log x)}
⇒ x