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Sagot :
To solve the differential equation
[tex]\[ y' + 9x e^y = 0, \][/tex]
we will proceed with the following steps:
1. Rewrite the equation: The given equation is
[tex]\[ y' + 9x e^y = 0. \][/tex]
First, let's isolate the derivative [tex]\( y' \)[/tex] (also written as [tex]\(\frac{dy}{dx}\)[/tex]):
[tex]\[ y' = -9x e^y. \][/tex]
2. Separate the variables: To solve this differential equation, we use the method of separation of variables. We will rewrite the equation so that all the [tex]\( y \)[/tex]-terms are on one side and all the [tex]\( x \)[/tex]-terms are on the other side:
[tex]\[ \frac{dy}{e^y} = -9x \, dx. \][/tex]
3. Integrate both sides: Now, we integrate both sides with respect to their respective variables. We start by integrating the left side with respect to [tex]\( y \)[/tex] and the right side with respect to [tex]\( x \)[/tex]:
[tex]\[ \int \frac{1}{e^y} \, dy = \int -9x \, dx. \][/tex]
The integral of [tex]\(\frac{1}{e^y}\)[/tex] with respect to [tex]\( y \)[/tex] is [tex]\( \int e^{-y} \, dy \)[/tex], which simplifies to:
[tex]\[ \int e^{-y} \, dy = -e^{-y}. \][/tex]
The integral of [tex]\(-9x\)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \int -9x \, dx = -9 \int x \, dx = -9 \left( \frac{x^2}{2} \right) = -\frac{9}{2} x^2. \][/tex]
4. Combine the integrals: Now we combine the results of the integrals:
[tex]\[ -e^{-y} = -\frac{9}{2} x^2 + C, \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
5. Solve for [tex]\( e^{-y} \)[/tex]: Multiply both sides of the equation by [tex]\(-1\)[/tex] to make the left side positive:
[tex]\[ e^{-y} = \frac{9}{2} x^2 - C. \][/tex]
6. Solve for [tex]\( y \)[/tex]: Take the natural logarithm of both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ -y = \ln\left( \frac{9}{2} x^2 - C \right). \][/tex]
Therefore:
[tex]\[ y = -\ln\left( \frac{9}{2} x^2 - C \right). \][/tex]
To simplify further, let [tex]\( C_1 = \frac{9}{2} x^2 - C \)[/tex], then we have:
[tex]\[ y = \ln\left( 1/C_1 \right) + \ln\left(2 \right). \][/tex]
Rewriting in a standard form, we get:
[tex]\[ \boxed{y(x) = \log \left(\frac{1}{C_1 + 9x^2} \right) + \log \left(2 \right)}. \][/tex]
So the final form of the solution to the differential equation is:
[tex]\[ y(x) = \log \left(\frac{1}{C_1 + 9x^2} \right) + \log \left(2 \right). \][/tex]
[tex]\[ y' + 9x e^y = 0, \][/tex]
we will proceed with the following steps:
1. Rewrite the equation: The given equation is
[tex]\[ y' + 9x e^y = 0. \][/tex]
First, let's isolate the derivative [tex]\( y' \)[/tex] (also written as [tex]\(\frac{dy}{dx}\)[/tex]):
[tex]\[ y' = -9x e^y. \][/tex]
2. Separate the variables: To solve this differential equation, we use the method of separation of variables. We will rewrite the equation so that all the [tex]\( y \)[/tex]-terms are on one side and all the [tex]\( x \)[/tex]-terms are on the other side:
[tex]\[ \frac{dy}{e^y} = -9x \, dx. \][/tex]
3. Integrate both sides: Now, we integrate both sides with respect to their respective variables. We start by integrating the left side with respect to [tex]\( y \)[/tex] and the right side with respect to [tex]\( x \)[/tex]:
[tex]\[ \int \frac{1}{e^y} \, dy = \int -9x \, dx. \][/tex]
The integral of [tex]\(\frac{1}{e^y}\)[/tex] with respect to [tex]\( y \)[/tex] is [tex]\( \int e^{-y} \, dy \)[/tex], which simplifies to:
[tex]\[ \int e^{-y} \, dy = -e^{-y}. \][/tex]
The integral of [tex]\(-9x\)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \int -9x \, dx = -9 \int x \, dx = -9 \left( \frac{x^2}{2} \right) = -\frac{9}{2} x^2. \][/tex]
4. Combine the integrals: Now we combine the results of the integrals:
[tex]\[ -e^{-y} = -\frac{9}{2} x^2 + C, \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
5. Solve for [tex]\( e^{-y} \)[/tex]: Multiply both sides of the equation by [tex]\(-1\)[/tex] to make the left side positive:
[tex]\[ e^{-y} = \frac{9}{2} x^2 - C. \][/tex]
6. Solve for [tex]\( y \)[/tex]: Take the natural logarithm of both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ -y = \ln\left( \frac{9}{2} x^2 - C \right). \][/tex]
Therefore:
[tex]\[ y = -\ln\left( \frac{9}{2} x^2 - C \right). \][/tex]
To simplify further, let [tex]\( C_1 = \frac{9}{2} x^2 - C \)[/tex], then we have:
[tex]\[ y = \ln\left( 1/C_1 \right) + \ln\left(2 \right). \][/tex]
Rewriting in a standard form, we get:
[tex]\[ \boxed{y(x) = \log \left(\frac{1}{C_1 + 9x^2} \right) + \log \left(2 \right)}. \][/tex]
So the final form of the solution to the differential equation is:
[tex]\[ y(x) = \log \left(\frac{1}{C_1 + 9x^2} \right) + \log \left(2 \right). \][/tex]
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