Math / Calculus

Question $y^{-}+2 x y-y^{4} e^{\left(3 x^{2}+2 x\right)^{11}}=0$

Studdy Solution

STEP 1

1. The given equation is a first-order differential equation.
2. The equation is linear in terms of $y'$ .
3. We will use an integrating factor to solve the differential equation.

STEP 2

1. Identify the standard form of the linear differential equation.
2. Determine the integrating factor.
3. Multiply through by the integrating factor.
4. Integrate both sides to find the solution.
5. Solve for $y$ .

STEP 3

First, identify the standard form of the linear differential equation. The standard form is:
$y' + P(x)y = Q(x)$
In the given equation:
$y' + 2xy - y^4 e^{(3x^2 + 2x)^{11}} = 0$
We can rewrite it as:
$y' + 2xy = y^4 e^{(3x^2 + 2x)^{11}}$
Here, $P(x) = 2x$ and $Q(x) = y^4 e^{(3x^2 + 2x)^{11}}$ .

STEP 4

Determine the integrating factor, $\mu(x)$ , which is given by:
$\mu(x) = e^{\int P(x) \, dx}$
Substitute $P(x) = 2x$ :
$\mu(x) = e^{\int 2x \, dx}$
$\mu(x) = e^{x^2}$

STEP 5

Multiply the entire differential equation by the integrating factor $\mu(x) = e^{x^2}$ :
$e^{x^2} y' + e^{x^2} (2xy) = e^{x^2} (y^4 e^{(3x^2 + 2x)^{11}})$
This simplifies to:
$\frac{d}{dx}(e^{x^2} y) = y^4 e^{(x^2 + (3x^2 + 2x)^{11})}$

STEP 6

Integrate both sides with respect to $x$ :
$\int \frac{d}{dx}(e^{x^2} y) \, dx = \int y^4 e^{(x^2 + (3x^2 + 2x)^{11})} \, dx$
The left side integrates to:
$e^{x^2} y = \int y^4 e^{(x^2 + (3x^2 + 2x)^{11})} \, dx + C$
where $C$ is the constant of integration.

STEP 7

Solve for $y$ :
$y = \frac{1}{e^{x^2}} \left( \int y^4 e^{(x^2 + (3x^2 + 2x)^{11})} \, dx + C \right)$
The solution for $y$ is given by the expression above, which involves an integral that may not have a simple closed form.

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