- #1

shamieh

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**implicit**form.

So I got:

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

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

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In summary: The key is that the integral constant is not just a constant, but a constant that may depend on the variable of integration. In this case, we have:\int \left( 2x - y \right) dx + g(y) = x^2 - xy + g(y)where $g(y)$ is a function of $y$. Differentiating this with respect to $y$ gives:2y - x = -x + g'(y)which means that $g'(y) = 2y$ and so $g(y) = y^2$. Substituting this back into the solution gives:x^2 - xy + y^2 = C

- #1

shamieh

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So I got:

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

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

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- #2

Prove It

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shamieh said:implicitform.

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.

- #3

shamieh

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- #4

cbarker1

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

- #5

Prove It

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shamieh said:uwhy 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 $\displaystyle \begin{align*} u = \frac{y}{x} \end{align*}$.

- #6

MarkFL

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

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

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

Hence, the solution is given implicitly by:

\(\displaystyle x^2-xy+y^2=C\)

Using the given initial conditions, we find:

\(\displaystyle 1^2-(1)(3)+3^2=C\)

\(\displaystyle C=7\)

Thus, the solution to the IVP is:

\(\displaystyle x^2-xy+y^2=7\)

- #7

shamieh

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MarkFL said: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. Thus we must have:exact

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

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

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

Hence, the solution is given implicitly by:

\(\displaystyle x^2-xy+y^2=C\)

Using the given initial conditions, we find:

\(\displaystyle 1^2-(1)(3)+3^2=C\)

\(\displaystyle C=7\)

Thus, the solution to the IVP is:

\(\displaystyle x^2-xy+y^2=7\)

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 $y$? $(2y-x)dy$ ?

- #8

MarkFL

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

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

- #9

shamieh

- 539

- 0

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

Last edited:

An IVP, or Initial Value Problem, is a mathematical equation that involves finding a function that satisfies a given differential equation, along with an initial condition. It is often used in scientific and engineering applications to model real-world situations.

Solving an IVP means finding a function that satisfies the given differential equation and initial condition. This involves finding a general solution to the differential equation and then using the initial condition to determine the specific solution.

Implicit form refers to an equation where the dependent variable is not explicitly written in terms of the independent variable. In other words, the solution is not expressed as a function of the independent variable, but rather as an equation involving both variables.

Leaving an IVP in implicit form allows for a more general solution to be found. It also allows for easier manipulation and further analysis of the equation. Additionally, some problems may only have implicit solutions and cannot be expressed explicitly.

Some common methods for solving IVPs in implicit form include using the separation of variables technique, the substitution method, and the integrating factor method. These methods involve manipulating the equation to isolate the dependent and independent variables and then using the initial condition to determine the specific solution.

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