What is the particular solution to the differential equation (dy)/(dx)=(x-3)y with the initial condition y(0)=2 ?

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What is the particular solution to the differential equation  (dy)/(dx)=(x-3)y with the initial condition  y(0)=2 ?

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Identify type of differential equation: Identify the type of differential equation. The given differential equation is a first-order linear differential equation in the form (dy)/(dx) + P(x)y = Q(x), where P(x) = -3 and Q(x) = x. This is also a separable differential equation since it can be written in the form (dy)/(y) = (x-3)dx.Separate variables: Separate the variables. We want to get all the y terms on one side and all the x terms on the other side. To do this, we divide both sides by y and multiply both sides by dx to get: (1/y)dy = (x-3)dxIntegrate both sides: Integrate both sides. We integrate the left side with respect to y and the right side with respect to x: ∫(1/y)dy = ∫(x-3)dx The integral of 1/y with respect to y is ln|y|, and the integral of (x-3) with respect to x is (x^2)/2 - 3x. So we have ln|y| = (x^2)/2 - 3x + C, where C is the constant of integration.Solve for constant of integration: Solve for the constant of integration using the initial condition. We are given the initial condition y(0) = 2. We plug x = 0 and y = 2 into the equation ln|y| = (x^2)/2 - 3x + C to find C: ln|2| = (0^2)/2 - 3(0) + C ln(2) = CWrite particular solution: Write the particular solution. Now that we have the constant C, we can write the particular solution: ln|y| = (x^2)/2 - 3x + ln(2) To solve for y, we exponentiate both sides to get rid of the natural logarithm: |y| = e^((x^2)/2 - 3x + ln(2)) Since y is positive for y(0) = 2, we can drop the absolute value: y = e^((x^2)/2 - 3x) * e^(ln(2)) y = 2e^((x^2)/2 - 3x)

What is the particular solution to the differential equation $\newline$ $\frac{dy}{dx}=(x-3)y$ with the initial condition $\newline$ $y(0)=2$ ?

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What is the particular solution to the differential equation \newlinedydx=(x−3)y\frac{dy}{dx}=(x-3)ydxdy​=(x−3)y with the initial condition \newliney(0)=2y(0)=2y(0)=2 ?

What is the particular solution to the differential equation $\newline$ $\frac{dy}{dx}=(x-3)y$ with the initial condition $\newline$ $y(0)=2$ ?