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Given:

[tex]\[2x + 2y \frac{dy}{dx} = 0\][/tex]

Solve for [tex]\(\frac{dy}{dx}\)[/tex].


Sagot :

To solve the differential equation [tex]\( 2x + 2y \frac{dy}{dx} = 0 \)[/tex], follow these steps:

1. Simplify the equation:
[tex]\[ 2x + 2y \frac{dy}{dx} = 0 \][/tex]
We can divide every term by 2 to simplify:
[tex]\[ x + y \frac{dy}{dx} = 0 \][/tex]

2. Rearrange the equation to separate the variables:
[tex]\[ y \frac{dy}{dx} = -x \][/tex]

3. Rewrite the equation in a separable form:
[tex]\[ y \frac{dy}{dx} = -x \][/tex]
Separate the variables [tex]\(y\)[/tex] and [tex]\(x\)[/tex]:
[tex]\[ y \, dy = -x \, dx \][/tex]

4. Integrate both sides to find the solution:
[tex]\[ \int y \, dy = \int -x \, dx \][/tex]
Perform the integration on both sides:
[tex]\[ \frac{y^2}{2} = -\frac{x^2}{2} + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.

5. Solve for [tex]\(y\)[/tex]:
Multiply both sides by 2 to simplify:
[tex]\[ y^2 = -x^2 + 2C \][/tex]
Rearrange it to find [tex]\(y\)[/tex]:
[tex]\[ y^2 = 2C - x^2 \][/tex]
Taking the square root of both sides:
[tex]\[ y = \pm \sqrt{2C - x^2} \][/tex]

6. Express the general form of the solution:
To present the solution in the desired form of [tex]\(f(x)\)[/tex], where the constant [tex]\(C\)[/tex] is replaced by a more general constant, say [tex]\(C_1\)[/tex]:
[tex]\[ y = \pm \sqrt{C_1 - x^2} \][/tex]
To eliminate the [tex]\( \pm \)[/tex] and to generally define it as a function, we note that it can be written as:
[tex]\[ y(x) = C_1 - \frac{x^2}{2} \][/tex]

Therefore, the general solution of the given differential equation [tex]\( 2x + 2y \frac{dy}{dx} = 0 \)[/tex] is:
[tex]\[ y(x) = C_1 - \frac{x^2}{2} \][/tex]