Solve the following differential equations (a)sin^(-1)(dy/dx)=x+y.
(b)[1+y^(2)]dx=[tan^(-1)y-x]dy?

Question

Isabella Ackerman · Accepted Answer

#"The gen. soln. is, "x+1=tan^-1y+c/e^(tan^-1y)#.

Part (a) : #sin^-1(dy/dx)=x+y. :. sin(x+y)=dy/dx#. Let #x+y=u. :. d/dx(x+y)=d/dx(u), i.e., 1+dy/dx=(du)/dx#. Subst.ing in the diff. eqn., # sinu=(du)/dx-1, or, (du)/dx=1+sinu#. #:. (du)/(1+sinu)=dx................"[Separable Variable]"#. #:. int(du)/(1+sinu)=intdx+c#. #:. int(1-sinu)/(1-sin^2u)du=x+c#, # or, int(1-sinu)/cos^2udu=x+c#. # :. int{1/cos^2u-sinu/cos^2u}du=x+c#. #:. tanu-secu=x+c#. Letting #u=x+y#, we get the gen. soln. as under : # tan(x+y)-sec(x+y)=x+c#. Part (b) : #(1+y^2)dx=(tan^-1y-x)dy#. #:. dx/dy=(tan^-1y-x)/(1+y^2)=tan^-1y/(1+y^2)-x/(1+y^2)#, # or, dx/dy+x*1/(1+y^2)=tan^-1y/(1+y^2)............(delta)#. This is a linear diff. eqn. of the form #dx/dy+x*p(y)=q(y)#. Let us find its Integrating Factor (IF), which is, #e^(p(y)dy)#. #intp(y)dy=int1/(1+y^2)dy=tan^-1y#. Multiplying #(delta)# by IF#=e^(tan^-1y)#, we get, #e^(tan^-1y)*dx/dy+x*e^(tan^-1y)/(1+y^2)=e^(tan^-1y)*tan^-1y/(1+y^2).# Note that, #d/dy{e^(tan^-1y)}=e^(tan^-1y)*1/(1+y^2)#. #:.e^(tan^-1y)*d/dy{x}+x*d/dy{e^(tan^-1y)}=e^(tan^-1y)*tan^-1y/(1+y^2)#. #:. d/dy{x*e^(tan^-1y)}=e^(tan^-1y)*tan^-1y/(1+y^2)#. # rArr x*e^(tan^-1y)=inte^(tan^-1y)*tan^-1y/(1+y^2)dy+c#, #=I+c,# where, #I=inte^(tan^-1y)*tan^-1y/(1+y^2)dy#, #=int(t*e^t)dt...[because, tan^-1y=t. :. 1/(1+y^2)dy=dt]#, #=t*inte^tdt-int{d/dt(t)*inte^tdt}dt......."[IBP, u=t, v'=e^t]"#, #=te^t-e^t#, #=(t-1)e^t#. # rArr I=(tan^-1y-1)e^(tan^-1y)#. Hence, #x*e^(tan^-1y)=(tan^-1y-1)e^(tan^-1y)+c, # # or, x+1=tan^-1y+c/e^(tan^-1y)#, is the desired gen. soln.!

Solve the following differential equations (a)sin^(-1)(dy/dx)=x+y. (b)[1+y^(2)]dx=[tan^(-1)y-x]dy?