What is a solution to the differential equation #dy/dx=x+y#?

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Answer 1 · 2016-10-19T21:11:58

#y = C_1e^x-x-1#

Let #u = x + y#

#=> (du)/dx = d/dx(x+y) = 1+dy/dx#

#=> dy/dx = (du)/dx-1#

Thus, making the substitutions into our original equation,

#(du)/dx-1 = u#

#=> (du)/(u+1) = dx#

#=> int(du)/(u+1)=intdx#

#=> ln(u+1) = x + C_0#

#=> e^(ln(u+1)) = e^(x+C_0)#

#=> u+1 = C_1e^x" "# (where #C_1 = e^(C_0)#)

Substituting #x+y = u# back in,

#=> x + y + 1 = C_1e^x#

#:. y = C_1e^x-x-1#