LỚP 10

test

To solve the differential equation \(y' + 2y = 3\), we can use the method of integrating factors.  First, we rewrite the equation in the standard form \(y' + P(x)y = Q(x)\), where \(P(x) = 2\) and \(Q(x) = 3\).  The integrating factor \(I(x)\) is given by \(I(x) = e^{\int P(x)dx}\). In this case, \(I(x) = e^{\int 2dx} = e^{2x}\).  Multiplying both sides of the differential equation by the integrating factor, we get \(e^{2x}y' + 2e^{2x}y = 3e^{2x}\).  Now, notice that the left side of the equation can be written as \((e^{2x}y)' = 3e^{2x}\).  Integrating both sides with respect to \(x\), we have \(\int (e^{2x}y)' dx = \int 3e^{2x} dx\).  Simplifying, we get \(e^{2x}y = \frac{3}{2}