Find the general solution of the following differential equation? #y' =sqrt(x+y)#

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Answer 1 · 2018-05-21T08:31:52

The general solution is #2(sqrt(x+y)-ln(sqrt(x+y)+1))=x+C#

The differential equation is

#dy/dx=sqrt(x+y)#

Perform the substitution

#v=x+y#

#(dv)/dx=1+dy/dx#

#dy/dx=(dv)/dx-1#

Therefore,

#(dv)/dx-1=sqrt(v)#

#(dv)/dx=sqrt(v)+1#

#(dv)/(sqrt(v)+1)=dx#

#int(dv)/(sqrt(v)+1)=intdx#

To integrate the LHS,

Let #u=sqrt(v)+1#, #=>#, #du=(dv)/(2sqrtv)#

Therefore,

#int(dv)/(sqrt(v)+1)=int(2sqrt(v)du)/(u)=2int((u-1)du)/(u)#

#=2int(1-1/u)du#

#=2u-2lnu#

#=2(sqrtv+1)-2ln(sqrtv+1)#

So,

#2(sqrtv+1)-2ln(sqrtv+1)=x+C#

#2(sqrt(x+y)-ln(sqrt(x+y)+1))=x+C#