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To determine the general solution of the differential equation
[tex]\[ \frac{dy}{dx} - 3y = 3x^2 e^x, \][/tex]
we will follow these steps:
1. Identify the type of differential equation: The given equation is a linear first-order differential equation.
2. Rewrite the equation in standard form: The standard form for a first-order linear differential equation is:
[tex]\[ \frac{dy}{dx} + P(x)y = Q(x), \][/tex]
where [tex]\( P(x) \)[/tex] and [tex]\( Q(x) \)[/tex] are functions of [tex]\( x \)[/tex]. In our equation,
[tex]\[ P(x) = -3 \quad \text{and} \quad Q(x) = 3x^2 e^x. \][/tex]
[tex]\[ \frac{dy}{dx} + (-3)y = 3x^2 e^x. \][/tex]
3. Determine the integrating factor [tex]\( \mu(x) \)[/tex]: The integrating factor is given by:
[tex]\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -3 \, dx} = e^{-3x}. \][/tex]
4. Multiply the entire equation by the integrating factor to make the left side a derivative of a product of functions:
[tex]\[ e^{-3x} \frac{dy}{dx} - 3e^{-3x} y = 3x^2 e^{x} e^{-3x}. \][/tex]
Simplify:
[tex]\[ e^{-3x} \frac{dy}{dx} - 3e^{-3x} y = 3x^2 e^{-2x}. \][/tex]
Recognize that the left side can be written as the derivative of [tex]\( y e^{-3x} \)[/tex]:
[tex]\[ \frac{d}{dx} (y e^{-3x}) = 3x^2 e^{-2x}. \][/tex]
5. Integrate both sides:
[tex]\[ \int \frac{d}{dx} (y e^{-3x}) \, dx = \int 3x^2 e^{-2x} \, dx. \][/tex]
On the left side, the integral of the derivative is simply [tex]\( y e^{-3x} \)[/tex], so:
[tex]\[ y e^{-3x} = \int 3x^2 e^{-2x} \, dx. \][/tex]
6. Evaluate the integral on the right side using integration by parts, where we let [tex]\( u = x^2 \)[/tex] and [tex]\( dv = 3e^{-2x} dx \)[/tex]. After performing integration by parts, we get:
[tex]\[ \int 3x^2 e^{-2x} \, dx = \int x^2 \cdot 3e^{-2x} \, dx. \][/tex]
The right side of our differential equation integrates to:
[tex]\[ \int 3x^2 e^{-2x} \, dx = -\frac{3}{2}x^2 e^{-2x} -\frac{3}{2}x e^{-2x} - \frac{3}{4}e^{-2x}. \][/tex]
Thus,
[tex]\[ y e^{-3x} = -\frac{3}{2}x^2 e^{-2x} -\frac{3}{2}x e^{-2x} - \frac{3}{4}e^{-2x} + C, \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
7. Solve for [tex]\( y \)[/tex] by multiplying through by [tex]\( e^{3x} \)[/tex]:
[tex]\[ y = e^{3x} \left(-\frac{3}{2}x^2 e^{-2x} -\frac{3}{2}x e^{-2x} - \frac{3}{4}e^{-2x} + C \right), \][/tex]
[tex]\[ y = e^{x} \left(-\frac{3}{2}x^2 -\frac{3}{2}x - \frac{3}{4} + C e^{2x} \right). \][/tex]
Simplify to obtain the general solution:
[tex]\[ y(x) = \left( C_1 e^{2x} - \frac{3}{2}x^2 - \frac{3}{2}x - \frac{3}{4} \right) e^{x}, \][/tex]
where [tex]\( C_1 \)[/tex] is an arbitrary constant of integration.
Thus, the general solution to the differential equation is:
[tex]\[ y(x) = \left( C_1 e^{2x} - \frac{3}{2}x^2 - \frac{3}{2}x - \frac{3}{4} \right) e^{x}. \][/tex]
[tex]\[ \frac{dy}{dx} - 3y = 3x^2 e^x, \][/tex]
we will follow these steps:
1. Identify the type of differential equation: The given equation is a linear first-order differential equation.
2. Rewrite the equation in standard form: The standard form for a first-order linear differential equation is:
[tex]\[ \frac{dy}{dx} + P(x)y = Q(x), \][/tex]
where [tex]\( P(x) \)[/tex] and [tex]\( Q(x) \)[/tex] are functions of [tex]\( x \)[/tex]. In our equation,
[tex]\[ P(x) = -3 \quad \text{and} \quad Q(x) = 3x^2 e^x. \][/tex]
[tex]\[ \frac{dy}{dx} + (-3)y = 3x^2 e^x. \][/tex]
3. Determine the integrating factor [tex]\( \mu(x) \)[/tex]: The integrating factor is given by:
[tex]\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -3 \, dx} = e^{-3x}. \][/tex]
4. Multiply the entire equation by the integrating factor to make the left side a derivative of a product of functions:
[tex]\[ e^{-3x} \frac{dy}{dx} - 3e^{-3x} y = 3x^2 e^{x} e^{-3x}. \][/tex]
Simplify:
[tex]\[ e^{-3x} \frac{dy}{dx} - 3e^{-3x} y = 3x^2 e^{-2x}. \][/tex]
Recognize that the left side can be written as the derivative of [tex]\( y e^{-3x} \)[/tex]:
[tex]\[ \frac{d}{dx} (y e^{-3x}) = 3x^2 e^{-2x}. \][/tex]
5. Integrate both sides:
[tex]\[ \int \frac{d}{dx} (y e^{-3x}) \, dx = \int 3x^2 e^{-2x} \, dx. \][/tex]
On the left side, the integral of the derivative is simply [tex]\( y e^{-3x} \)[/tex], so:
[tex]\[ y e^{-3x} = \int 3x^2 e^{-2x} \, dx. \][/tex]
6. Evaluate the integral on the right side using integration by parts, where we let [tex]\( u = x^2 \)[/tex] and [tex]\( dv = 3e^{-2x} dx \)[/tex]. After performing integration by parts, we get:
[tex]\[ \int 3x^2 e^{-2x} \, dx = \int x^2 \cdot 3e^{-2x} \, dx. \][/tex]
The right side of our differential equation integrates to:
[tex]\[ \int 3x^2 e^{-2x} \, dx = -\frac{3}{2}x^2 e^{-2x} -\frac{3}{2}x e^{-2x} - \frac{3}{4}e^{-2x}. \][/tex]
Thus,
[tex]\[ y e^{-3x} = -\frac{3}{2}x^2 e^{-2x} -\frac{3}{2}x e^{-2x} - \frac{3}{4}e^{-2x} + C, \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
7. Solve for [tex]\( y \)[/tex] by multiplying through by [tex]\( e^{3x} \)[/tex]:
[tex]\[ y = e^{3x} \left(-\frac{3}{2}x^2 e^{-2x} -\frac{3}{2}x e^{-2x} - \frac{3}{4}e^{-2x} + C \right), \][/tex]
[tex]\[ y = e^{x} \left(-\frac{3}{2}x^2 -\frac{3}{2}x - \frac{3}{4} + C e^{2x} \right). \][/tex]
Simplify to obtain the general solution:
[tex]\[ y(x) = \left( C_1 e^{2x} - \frac{3}{2}x^2 - \frac{3}{2}x - \frac{3}{4} \right) e^{x}, \][/tex]
where [tex]\( C_1 \)[/tex] is an arbitrary constant of integration.
Thus, the general solution to the differential equation is:
[tex]\[ y(x) = \left( C_1 e^{2x} - \frac{3}{2}x^2 - \frac{3}{2}x - \frac{3}{4} \right) e^{x}. \][/tex]
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