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The solution of the differential equation is :
dy/dx + 3y = 2e^{3}x
p = 3, q = 2e^{3}x
∫p.dx = 3x
Integrating factor (I.F) = e^{3}x
y(I.F) = ∫Q(I.F) dx
ye^{3}x = ∫2e^{3}x e^{3}x dx
ye^{3}x = ∫2e^{6}x dx
ye^{3}x = 2∫e^{6}x dx
ye^{3}x = 2/6[e^{6}x] + c
ye^{3}x = ⅓[e^{6}x] + c
Dividing by e^{3}x, we get
y = ⅓[e^{3}x] + ce^{(3x)}
The integrating factor of differential equation is :
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
The solution of the differential equation x dy = (2y + 2x^{4} + x^{2}) dx is:
xdy = (2y + 2x^{4 }+ x^{2})dx
→ dy/dx − (2x)y = 2x^{3} + 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/x^{2}) = ∫1/x^{2}(2x^{3} + x)dx
=∫(2x + 1/x)dx
= x^{2} + lnx + C
⟹ y = x^{4} + x^{2}lnx + c
The integrating factor of differential equation is :
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)}
=sec^{2}x
The solution of the differential equation is :
dy/dx + y/x = x^{2}
differential equation is in the form : dy/dx + Py = Q
P = 1/x Q = x^{2}
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 = ∫x^{2} * x * dx + c
yx = ∫x^{3} dx + c
xy = [x^{4}]/4 + c
The solution of the differential equation is :
The solution of the differential equation is :
The given differential equation may be written as
dy/dx + (2x/(x^{2}+1)y = √x2+4/(x^{2}+1) ... (i)
This is of the form dy/dx + Py = Q,
where P=2x/(x^{2}+1) and Q=(√x^{2}+4)/(x^{2}+1)
Thus, the given differential equation is linear.
IF=e^{(∫Pdx)}
= e(∫2x(x^{2}+1)dx)
= e(log(x^{2}+1) = (x^{2}+1)
So, the required solution is given by
y × IF = ∫{Q×IF}dx + C,
i.e., y(x^{2}+1)=∫(√x^{2}+4)/(x^{2}+1)×(x^{2}+1)dx
⇒y(x^{2}+1)=∫(√x^{2}+4)dx
=1/2x (√x^{2}+4) +1/2 × (2)^{2} × logx+(√x^{2}+4) + C
=1/2x (√x^{2}+4) + 2logx+(√x^{2}+4) + C.
Hence, y(x^{2}+1) = 1/2x(√x^{2}+4) + 2logx+√x^{2}+4 + C is the required solution.
The integrating factor of differential equation is :
The integrating factor of differential equation is :
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
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