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 $du=2xcolor(white).dx$ :

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

$y=5/2int1/sqrtudu$

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

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

$y=5sqrtu+C$

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

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