MHB Solving the IVP, leaving it in Implicit Form

[shamieh]

Solve the IVP. $(2x-y)dx + (2y-x)dy = 0 $. $y(1) = 3$. Leave solution in implicit form.

So I got:

$\frac{dy}{dx} = \frac{-(2x-y)}{2y-x}$

Would this be correct since I didn't explicitly solve for $dy$ ?

 

[Prove It]

Solve the IVP. $(2x-y)dx + (2y-x)dy = 0 $. $y(1) = 3$. Leave solution in implicit form.

So I got:

$\frac{dy}{dx} = \frac{-(2x-y)}{2y-x}$

Would this be correct since I didn't explicitly solve for $dy$ ?

Yes, now rewrite it as

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{y - 2x}{2y - x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{x \, \left( \frac{y}{x} - 2 \right) }{x \, \left( 2\,\frac{y}{x} - 1 \right) } \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\frac{y}{x} - 2}{2\,\frac{y}{x} - 1} \end{align*}$

Now substitute $\displaystyle \begin{align*} u = \frac{y}{x} \implies y = u\,x \implies \frac{\mathrm{d}y}{\mathrm{d}x} = u + x\,\frac{\mathrm{d}u}{\mathrm{d}x} \end{align*}$ and the DE becomes

$\displaystyle \begin{align*} u + x\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{u - 2}{2u - 1} \end{align*}$

which is now a separable equation.

 

[shamieh]

Why are you substituting in u why can't I just leave it in the form as my original answer is above? Why would that be incorrect?

 

[cbarker1]

First, it needs to be integrated. Then Substitute back u to x and y. then it would be implicit form.

 

[Prove It]

Why are you substituting in u why can't I just leave it in the form as my original answer is above? Why would that be incorrect?

Well, you tell me how you would separate the variables? Or how you could write it in some other form where a solution can be found.

Generally speaking, if it's possible to write your DE as $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = f \left( \frac{y}{x} \right) \end{align*}$ it can be turned into a separable equation with the substitution $\displaystyle \begin{align*} u = \frac{y}{x} \end{align*}$.

 

[MarkFL]

While Prove It's suggestion leads to a homogeneous first order equation, for which his suggested substitution leads to a separable equation, you could also observe that the ODE in its original form is exact. Thus we must have:

$$F(x,y)=\int 2x-y\,dx+g(y)=x^2-xy+g(y)$$

Differentiating this with respect to $y$, we find:

$$2y-x=-x+g'(y)\implies g(y)=y^2$$

Hence, the solution is given implicitly by:

$$x^2-xy+y^2=C$$

Using the given initial conditions, we find:

$$1^2-(1)(3)+3^2=C$$

$$C=7$$

Thus, the solution to the IVP is:

$$x^2-xy+y^2=7$$

 

[shamieh]

While Prove It's suggestion leads to a homogeneous first order equation, for which his suggested substitution leads to a separable equation, you could also observe that the ODE in its original form is exact. Thus we must have:

$$F(x,y)=\int 2x-y\,dx+g(y)=x^2-xy+g(y)$$

Differentiating this with respect to $y$, we find:

$$2y-x=-x+g'(y)\implies g(y)=y^2$$

Hence, the solution is given implicitly by:

$$x^2-xy+y^2=C$$

Using the given initial conditions, we find:

$$1^2-(1)(3)+3^2=C$$

$$C=7$$

Thus, the solution to the IVP is:

$$x^2-xy+y^2=7$$

you're saying first take the integral w.r.t $x$

$\int 2x-y dx + g(y) = x^2 - xy + g(y)$ right?

then you're saying differentiate "this" with respect to $y$. I'm lost what are we differentiating w.r.t $y$? $(2y-x)dy$ ?

 

[MarkFL]

Since the equation is exact, we know that:

$$\pd{F}{y}=N(x,y)=2y-x$$

Your textbook should have a section on exact equations that explains the method in more detail. :D

 

[shamieh]

I understand now... g'(y) implies that g(y) must be the A.D. of g'(y) which is $y^2$