What is a general solution to the differential equation #y'+2xy=2x^3#?

2 Answers
Jul 26, 2016

The Gen. Soln. is,

#ye^(x^2)=(x^2-1)e^(x^2)+C#, or, #y=x^2-1+Ce^(-x^2)#.

Explanation:

This is a Linear Diff. Eqn. # : dy/dx+P(x)*y=Q(x)...............(1)#,

where, #P(x)# and #Q(x)# are funs. of variable #x#.

To find its Gen. Soln. , we have to multiply it by, Integrating

Factor given by, #I.F.=e^(intP(x)dx)#.

Comparing the given diff. eqn. with #(1)#, we have,

#P(x)=2xrArrintP(x)dx=int2xdx=x^2rArr I.F.=e^(x^2)#.

Multiplying the eqn. throughout by #I,F.#, we get,

#e^(x^2)dy/dx+2xye^(x^2)=2x^3e^(x^2)#

#:. e^(x^2)d/dx(y)+yd/dx(e^(x^2))=2x^3e^(x^2)#.

#:. d/dx(y*e^(x^2))=2x^3e^(x^2)#

#:. y*e^(x^2)=int2x^3e^(x^2)dx............(2)#

To evaluate the integral, say #I#, on #R.H.S.#, we will subst.,

#x^2=t#, so that, #2xdx=dt#.

#:.I=intte^tdt#

#=tinte^tdt-int[(d/dt(t)*inte^tdt]dt............[IBP]#

#=te^t-inte^tdt=te^t-e^t=(t-1)e^t=(x^2-1)e^(x^2)+C#.

By #(2)#, then, the Gen. Soln. is,

#ye^(x^2)=(x^2-1)e^(x^2)+C#, or, #y=x^2-1+Ce^(-x^2)#.

Jul 26, 2016

#y = x^2 - 1 + C e^(-x^2)#

Explanation:

#y'+2xy=2x^3#

this is not separable. we can use an integrating factor # eta(x)#

here #eta = exp (int 2x \ dx) = e^(x^2) #

so
#e^(x^2) y'+e^(x^2) 2xy=e^(x^2) 2x^3#

or

#d/dx (e^(x^2) y)=e^(x^2) 2x^3#

#e^(x^2) y = int \ e^(x^2) 2x^3 dx#

#e^(x^2) y = int \ d/dx ( e^(x^2) ) x^2 dx#

using IBP

#e^(x^2) y = x^2 e^(x^2) - int \ e^(x^2) d/dx( x^2) dx#

#e^(x^2) y = x^2 e^(x^2) - 2 int \ x e^(x^2) dx#

and spotting the pattern in the integrand

#e^(x^2) y = x^2 e^(x^2) - 2 int \ d/dx(1/2 e^(x^2)) dx#

#e^(x^2) y = x^2 e^(x^2) - e^(x^2) + C#

#y = x^2 - 1 + C e^(-x^2)#