find_the_fun
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I'm unclear how to do these kinds of questions. Here's an example
Determine whether the given first-order differential equation is linear in the indicated dependent variable by matching it with the equation $$a_1(x)\frac{dy}{dx} = + a_o(x)y=g(x)$$
[math](y^2-1)dx+xdy=0[/math]; 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.
[math](y^2-1)dx+xdy=0[/math]
[math](y^2-1)+x\frac{dy}{dx}=0[/math]
[math]x\frac{dy}{dx}+y^2=1[/math]
Therefore [math]a_1(x)=x[/math], [math]a_o(x)=invalid[/math] and [math]g(x)=1[/math] Since one is invalid the equation is non linear in y.
For it being linear in x
[math](y^2-1)dx+xdy=0[/math]
[math](y^2-1)\frac{dx}{dy}+x=0[/math]
Determine whether the given first-order differential equation is linear in the indicated dependent variable by matching it with the equation $$a_1(x)\frac{dy}{dx} = + a_o(x)y=g(x)$$
[math](y^2-1)dx+xdy=0[/math]; 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.
[math](y^2-1)dx+xdy=0[/math]
[math](y^2-1)+x\frac{dy}{dx}=0[/math]
[math]x\frac{dy}{dx}+y^2=1[/math]
Therefore [math]a_1(x)=x[/math], [math]a_o(x)=invalid[/math] and [math]g(x)=1[/math] Since one is invalid the equation is non linear in y.
For it being linear in x
[math](y^2-1)dx+xdy=0[/math]
[math](y^2-1)\frac{dx}{dy}+x=0[/math]