Obtain the general solution of ODE : 5xy' + 2y = 6x

To obtain the general solution of the given ordinary differential equation (ODE), we will follow these steps:

Step 1: Identify the type of ODE The given equation is a first-order linear ODE, which can be written in the form: y' + P(x)y = Q(x), where P(x) = 2/x and Q(x) = 6x/5.

Step 2: Find the integrating factor The integrating factor (IF) is given by the formula: IF = e^(∫P(x)dx). In this case, P(x) = 2/x, so the integrating factor is IF = e^(∫2/x dx).

Step 3: Evaluate the integral To evaluate the integral, we can rewrite it as: ∫2/x dx = 2∫(1/x) dx = 2ln|x| + C, where C is the constant of integration.

Step 4: Calculate the integrating factor Using the evaluated integral, the integrating factor becomes: IF = e^(2ln|x| + C) = e^(ln|x|^2 + C) = e^(ln|x|^2) * e^C = |x|^2 * e^C.

Step 5: Multiply the ODE by the integrating factor Multiplying the given ODE by the integrating factor, we get: |x|^2 * e^C * 5xy' + |x|^2 * e^C * 2y = |x|^2 * e^C * 6x.

Step 6: Simplify the equation Simplifying the equation, we have: 5x^3y' + 2x^2y = 6x^3 * e^C.

Step 7: Integrate both sides of the equation Integrating both sides of the equation, we obtain: ∫(5x^3y' + 2x^2y) dx = ∫(6x^3 * e^C) dx.

Step 8: Evaluate the integrals Evaluating the integrals, we get