# Solve dy dx + 1 2 y = 3 2 , with y(0) = 2?…

To solve the given differential equation, we will use the method of integrating factors. The equation is a linear first-order ordinary differential equation of the form dy/dx + P(x)y = Q(x), where P(x) = 1/2 and Q(x) = 3/2.

The integrating factor for this equation is defined as μ(x) = e^(∫P(x)dx). In our case, P(x) = 1/2, so we need to find ∫(1/2)dx. Integrating 1/2 with respect to x gives us (1/2)x. Therefore, the integrating factor μ(x) = e^((1/2)x).

Next, we multiply both sides of the differential equation by the integrating factor μ(x) to obtain:

e^((1/2)x)dy/dx + (1/2)e^((1/2)x)y = (3/2)e^((1/2)x).

Now, we can rewrite the left side of the equation by applying the product rule of differentiation:

d/dx(ye^(1/2x)) = (3/2)e^((1/2)x).

To find y, we need to integrate both sides of the equation with respect to x. Integrating (3/2)e^((1/2)x)dx gives us:

∫((3/2)e^((1/2)x))dx = (3/2)∫e^((1/2)x)dx.

Using the substitution method, let u = (1/2)x, so du = (1/2)dx. We can rewrite the integral as:

(3/2)∫2e^udu.

Simplifying, we have:

(3/2)∫e^udu = (3/2)e^u + C,

where C is the constant of integration. Substituting back for u, we get:

(3/2)e^u = (3/2)e^((1/2)x) + C.

Let’s call the constant of integration C_1, so we have:

(3/2)e^u = (3/2)e^((1/2)x) + C_1.

Now, we can solve for y by dividing both sides of the equation by the integrating factor μ(x) = e^((1/2)x):

ye^(1/2x) = e^((1/2)x) + (2/3)C_1.

Next, we can isolate y by multiplying both sides by e^(-1/2x):

y = e^((1/2)x) + (2/3)C_1)e^(-1/2x).

To obtain the particular solution, we need to consider the initial condition y(0) = 2. Substituting x = 0 into the equation, we have:

2 = e^(0) + (2/3)C_1e^(0),
2 = 1 + (2/3)C_1.

Simplifying the equation, we find:

(2/3)C_1 = 1,

C_1 = (3/2).

Substituting this value back into the equation for y, we obtain:

y = e^((1/2)x) + (2/3)(3/2)e^(-1/2x),
y = e^((1/2)x) + (1/2)e^(-1/2x).

Therefore, the solution to the given differential equation is:

y = e^((1/2)x) + (1/2)e^(-1/2x).