solve the initial value problem y'=(7xy)/(7x^2+7y^2), y(1)=1. Express your answer in the form F(x,y)=7/14, where F(x,y) =? Hint: This is a homogeneous equation. to solve use the substitution y=xu, it will transform into separable differential equation in u


We are tasked with solving the initial value problem:

\[
y' = \frac{7xy}{7x^2 + 7y^2}, \quad y(1) = 1
\]

### Step 1: Simplify the given differential equation
First, simplify the equation by canceling out common factors of 7:

\[
y' = \frac{xy}{x^2 + y^2}
\]

### Step 2: Use the substitution \( y = xu \)
According to the hint, we substitute \( y = xu \), where \( u \) is a function of \( x \). This substitution transforms the original equation into a more solvable form.

- Differentiate \( y = xu \) with respect to \( x \):

\[
y' = u + x \frac{du}{dx}
\]

### Step 3: Substitute into the original equation
Substitute \( y = xu \) and \( y' = u + x \frac{du}{dx} \) into the original equation \( y' = \frac{xy}{x^2 + y^2} \):

\[
u + x \frac{du}{dx} = \frac{x(xu)}{x^2 + (xu)^2}
\]

Simplifying the right-hand side:

\[
u + x \frac{du}{dx} = \frac{x^2 u}{x^2(1 + u^2)} = \frac{u}{1 + u^2}
\]

Now the equation becomes:

\[
u + x \frac{du}{dx} = \frac{u}{1 + u^2}
\]

### Step 4: Rearrange into a separable form
Move all terms involving \( u \) to one side:

\[
x \frac{du}{dx} = \frac{u}{1 + u^2} - u
\]

Simplify the right-hand side:

\[
x \frac{du}{dx} = u \left( \frac{1}{1 + u^2} - 1 \right)
\]

This simplifies further to:

\[
x \frac{du}{dx} = \frac{-u^3}{1 + u^2}
\]

### Step 5: Separate variables
Now, we separate the variables \( u \) and \( x \):

\[
\frac{1 + u^2}{u^3} du = -\frac{dx}{x}
\]

Simplify the left-hand side:

\[
\left( \frac{1}{u^2} + \frac{1}{u^4} \right) du = -\frac{dx}{x}
\]

### Step 6: Integrate both sides
Now, integrate both sides:

\[
\int \left( \frac{1}{u^2} + \frac{1}{u^4} \right) du = -\int \frac{dx}{x}
\]

The integrals are:

\[
-\frac{1}{u} - \frac{1}{3u^3} = -\ln |x| + C
\]

### Step 7: Use initial condition \( y(1) = 1 \)
Recall that \( y = xu \), and at \( x = 1 \), \( y = 1 \). This gives:

\[
1 = 1 \cdot u \quad \Rightarrow \quad u = 1
\]

Substitute \( u = 1 \) and \( x = 1 \) into the equation:

\[
-\frac{1}{1} - \frac{1}{3(1)^3} = -\ln |1| + C
\]

Simplifying:

\[
-1 - \frac{1}{3} = 0 + C \quad \Rightarrow \quad C = -\frac{4}{3}
\]

### Step 8: Write the general solution
The general solution is:

\[
-\frac{1}{u} - \frac{1}{3u^3} = -\ln |x| - \frac{4}{3}
\]

Multiplying by -1 and rearranging:

\[
\frac{1}{u} + \frac{1}{3u^3} = \ln |x| + \frac{4}{3}
\]

Finally, substitute back \( u = \frac{y}{x} \):

\[
\frac{x}{y} + \frac{x^3}{3y^3} = \ln |x| + \frac{4}{3}
\]

This is the implicit solution for the initial value problem.

### Expressing in the form \( F(x, y) = \frac{7}{14} \)
In this case, the function \( F(x, y) \) corresponds to:

\[
F(x, y) = \frac{x}{y} + \frac{x^3}{3y^3} - \ln |x| - \frac{4}{3}
\]

This expression implicitly satisfies the differential equation.

Would you like more details on any specific step, or help understanding this solution?

### Follow-up Questions:
1. How do you verify that this is a homogeneous equation?
2. Can you explain why the substitution \( y = xu \) works in simplifying homogeneous equations?
3. What is the method of separation of variables, and why is it useful here?
4. How does the initial condition help in solving for the constant \( C \)?
5. Can this differential equation be solved using another method besides substitution?

### Tip:
Always check if an equation is homogeneous by analyzing the degree of each term before applying substitutions.

solve the initial value problem y'=(7xy)/(7x^2+7y^2), y(1)=1. Express your answer in the form F(x,y)=7/14, where F(x,y) =? Hint: This is a homogeneous equation. To solve, use the substitution y=xu; it will transform into a separable differential equation in u.

Explore how to solve the initial value problem y' = (7xy)/(7x^2 + 7y^2), y(1)=1. Learn the substitution y=xu to transform the equation into a separable differential form, utilizing key concepts of homogeneous differential equations and separation of variables.

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