Question

(dy/dx)-(2y/x)=5x^2/(2+x)(3-2x)​

Answer 1

Answer:

y = 5x*2/7 log(x+2/3-2x)c

Answer 2

The general solution to the differential equation will be y = (x^2-6x+5)[(2+x)/5]^1 + [(3-2x)/5]^4 + Cx^2.

Given: The differential equation (dy/dx) - (2y/x) = 5x^2/(2+x)(3-2x)

To Find: the general solution of differential equation:

Solution:

(dy/dx) - (2y/x) = 5x^2/(2+x)(3-2x)

Let us use the method integrating of factors.

Initially, we need to find the integrating factor, which is given by:

IF(x) = e^(∫-2/x dx)

= e^(-2ln|x|)

= 1/x^2

Now let us multiply both sides of the differential equation by the integrating factor, then:

1/x^2(dy/dx) - 2/x^3(y) = 5x^2/(x^2+5x-6)

We  can  simplify the LHS using the product rule:

d/dx[1/x^2(y)] = -2/x^3(y) + (dy/dx)(-2/x^3)

So we have:d/dx[1/x^2(y)] = 5x^2/(x^2+5x-6)

Let us integrate both sides with x then:

1/x^2(y) = ∫5x^2/(x^2+5x-6) dx

To evaluate the integral on the right-hand side, we use partial fraction decomposition:

5x^2/(x^2+5x-6) = A/(2+x) + B/(3-2x)

Multiplying both sides by (2+x)(3-2x), we get:

5x^2 = A(3-2x) + B(2+x)

Solving for A and B, we get:

A = 1/5, B = 4/5

Substituting back into the partial fraction decomposition, we have:

5x^2/(x^2+5x-6) = 1/5(2+x) + 4/5(3-2x)

Integrating both sides, we get:

ln|x^2-6x+5|/x^2 = (1/5)ln|2+x| - (4/5)ln|3-2x| + C

where C is the constant of integration.

To simplifying, we have:

y = (x^2-6x+5)[(2+x)/5]^1 + [(3-2x)/5]^4 + Cx^2

Hence, the general solution to the differential equation will be

y = (x^2-6x+5)[(2+x)/5]^1 + [(3-2x)/5]^4 + Cx^2, where C is an arbitrary constant.

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