How to you find the general solution of #sqrt(x^2-9)y'=5x#?

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Answer 1 · 2017-01-18T21:13:40

#y=5sqrt(x^2-9)+C#

Writing #y'# as #dy/dx#:

#sqrt(x^2-9)dy/dx=5x#

Then:

#dy/dx=(5x)/sqrt(x^2-9)#

Separating the variables by treating #dy/dx# as a quotient then integrating both sides:

#intdy=int(5x)/sqrt(x^2-9)dx#

#y=int(5x)/sqrt(x^2-9)dx#

Solve the remaining integral by letting #u=x^2-9#, implying that #du=2xcolor(white).dx#:

#y=5/2int(2x)/sqrt(x^2-9)dx#

#y=5/2int1/sqrtudu#

#y=5/2intu^(-1/2)du#

#y=5/2(u^(1/2)/(1/2))+C#

#y=5sqrtu+C#

#y=5sqrt(x^2-9)+C#