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Solve the following differential equation ; (3x^2 + 6xy^2) dx + (6x^2 + 4y^2) dy = 0Asked by Hakizimana
Solve the following differential equation ;
(3x^2 + 6xy^2) dx + ((6x^2)y + 4y^2) dy = 0
(3x^2 + 6xy^2) dx + ((6x^2)y + 4y^2) dy = 0
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Answered by
GPT-4o mini
To solve the differential equation
\[
(3x^2 + 6xy^2) \, dx + ((6x^2)y + 4y^2) \, dy = 0,
\]
we can rewrite it in the form
\[
M(x, y) \, dx + N(x, y) \, dy = 0,
\]
where
\[
M(x, y) = 3x^2 + 6xy^2
\]
and
\[
N(x, y) = (6x^2y + 4y^2).
\]
To check if the equation is exact, we need to verify if
\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.
\]
First, we compute the partial derivatives:
1. Compute \(\frac{\partial M}{\partial y}\):
\[
M(x, y) = 3x^2 + 6xy^2 \implies \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2) + \frac{\partial}{\partial y}(6xy^2) = 0 + 12xy = 12xy.
\]
2. Compute \(\frac{\partial N}{\partial x}\):
\[
N(x, y) = 6x^2y + 4y^2 \implies \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(6x^2y) + \frac{\partial}{\partial x}(4y^2) = 12xy + 0 = 12xy.
\]
Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 12xy\), the equation is exact.
Next, we find a function \(\Psi(x, y)\) such that
\[
\frac{\partial \Psi}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial \Psi}{\partial y} = N(x, y).
\]
### Step 1: Integrate \(M\) with respect to \(x\):
\[
\Psi(x, y) = \int (3x^2 + 6xy^2) \, dx = x^3 + 3xy^2 + h(y),
\]
where \(h(y)\) is an arbitrary function of \(y\).
### Step 2: Differentiate \(\Psi\) with respect to \(y\):
\[
\frac{\partial \Psi}{\partial y} = \frac{\partial}{\partial y}(x^3 + 3xy^2 + h(y)) = 6xy + h'(y).
\]
### Step 3: Set this equal to \(N(x,y)\):
\[
6xy + h'(y) = 6x^2y + 4y^2.
\]
From this, we obtain:
\[
h'(y) = 4y^2.
\]
### Step 4: Integrate \(h'(y)\) with respect to \(y\):
\[
h(y) = \int 4y^2 \, dy = \frac{4y^3}{3} + C,
\]
where \(C\) is a constant.
### Step 5: Substitute \(h(y)\) back into \(\Psi\):
\[
\Psi(x, y) = x^3 + 3xy^2 + \frac{4y^3}{3} + C.
\]
### Conclusion
Setting \(\Psi(x, y) = k\) (where \(k\) is a constant), we have the implicit solution to the differential equation:
\[
x^3 + 3xy^2 + \frac{4y^3}{3} = C.
\]
This represents the family of solutions to the given differential equation.
\[
(3x^2 + 6xy^2) \, dx + ((6x^2)y + 4y^2) \, dy = 0,
\]
we can rewrite it in the form
\[
M(x, y) \, dx + N(x, y) \, dy = 0,
\]
where
\[
M(x, y) = 3x^2 + 6xy^2
\]
and
\[
N(x, y) = (6x^2y + 4y^2).
\]
To check if the equation is exact, we need to verify if
\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.
\]
First, we compute the partial derivatives:
1. Compute \(\frac{\partial M}{\partial y}\):
\[
M(x, y) = 3x^2 + 6xy^2 \implies \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2) + \frac{\partial}{\partial y}(6xy^2) = 0 + 12xy = 12xy.
\]
2. Compute \(\frac{\partial N}{\partial x}\):
\[
N(x, y) = 6x^2y + 4y^2 \implies \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(6x^2y) + \frac{\partial}{\partial x}(4y^2) = 12xy + 0 = 12xy.
\]
Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 12xy\), the equation is exact.
Next, we find a function \(\Psi(x, y)\) such that
\[
\frac{\partial \Psi}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial \Psi}{\partial y} = N(x, y).
\]
### Step 1: Integrate \(M\) with respect to \(x\):
\[
\Psi(x, y) = \int (3x^2 + 6xy^2) \, dx = x^3 + 3xy^2 + h(y),
\]
where \(h(y)\) is an arbitrary function of \(y\).
### Step 2: Differentiate \(\Psi\) with respect to \(y\):
\[
\frac{\partial \Psi}{\partial y} = \frac{\partial}{\partial y}(x^3 + 3xy^2 + h(y)) = 6xy + h'(y).
\]
### Step 3: Set this equal to \(N(x,y)\):
\[
6xy + h'(y) = 6x^2y + 4y^2.
\]
From this, we obtain:
\[
h'(y) = 4y^2.
\]
### Step 4: Integrate \(h'(y)\) with respect to \(y\):
\[
h(y) = \int 4y^2 \, dy = \frac{4y^3}{3} + C,
\]
where \(C\) is a constant.
### Step 5: Substitute \(h(y)\) back into \(\Psi\):
\[
\Psi(x, y) = x^3 + 3xy^2 + \frac{4y^3}{3} + C.
\]
### Conclusion
Setting \(\Psi(x, y) = k\) (where \(k\) is a constant), we have the implicit solution to the differential equation:
\[
x^3 + 3xy^2 + \frac{4y^3}{3} = C.
\]
This represents the family of solutions to the given differential equation.
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