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#### find_the_fun

##### Active member

- Feb 1, 2012

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Determine whether the given first-order differential equation is linear in the indicated dependent variable by matching it with the equation \(\displaystyle a_1(x)\frac{dy}{dx} = + a_o(x)y=g(x)\)

\(\displaystyle (y^2-1)dx+xdy=0\); in y; in x

The answer is it's linear in x but nonlinear in y.

First off I don't understand "indicated dependent variable", is it saying first consider y as a function of x and then consider x as a function of y? What difference does this have on the solution?

Here's what I tried.

\(\displaystyle (y^2-1)dx+xdy=0\)

\(\displaystyle (y^2-1)+x\frac{dy}{dx}=0\)

\(\displaystyle x\frac{dy}{dx}+y^2=1\)

Therefore \(\displaystyle a_1(x)=x\), \(\displaystyle a_o(x)=invalid\) and \(\displaystyle g(x)=1\) Since one is invalid the equation is non linear in y.

For it being linear in x

\(\displaystyle (y^2-1)dx+xdy=0\)

\(\displaystyle (y^2-1)\frac{dx}{dy}+x=0\)