Differential Equations Test - 9

dy/dx + 3y = 2e³ x
p = 3,   q = 2e³ x
∫p.dx = 3x
Integrating factor (I.F) = e³ x
y(I.F) = ∫Q(I.F) dx
ye³ x = ∫2e³ x e³ x dx
ye³ x = ∫2e⁶ x dx
ye³ x = 2 ∫e⁶ x dx
ye³ x = 2/6[e⁶ x] + c
ye³ x = ⅓[e⁶ x] + c
Dividing by e³ x, we get
 y = ⅓[e³ x] + 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 + x² lnx + 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)}
=sec² x

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