y′=(2y+x)/x, y(1)=4
1. The resulting differential equation in x and u can be written as xu'=------
2.Separating variables we arrive at -----du= dx/x
3.Integrating both sides and simplifying, the solution can be written in the form u+1=Cf(x)
where C is an arbitrary constant and f(x)=-----
4.Transforming back into the variables x and y and using the initial condition to find C we find the explicit solution of the initial value problem is y=------
1. To write the resulting differential equation in terms of x and u, we can rewrite the given equation y' = (2y + x)/x as xu' = 2u + 1. Here, we replace y' with dy/dx and let u = y/x.
2. To separate the variables, we can rearrange the equation xu' = 2u + 1 as (1/u + 1) du = dx/x. Here, we divide both sides by (2u + 1) to obtain the fraction on the left.
3. Integrating both sides of the separated equation gives us ∫(1/u + 1) du = ∫dx/x. The integral of (1/u + 1) du can be simplified to ln|u| + u + C, where C is a constant. The integral of dx/x is ln|x| + C1, where C1 is another constant.
Combining these results, we have ln|u| + u + C = ln|x| + C1. We can rewrite this equation as u + 1 = Cf(x), where C = e^(C1) is an arbitrary constant and f(x) = ln|x|.
4. Transforming back into the variables x and y, we replace u with y/x in the equation u + 1 = Cf(x). This gives us y/x + 1 = Cf(x). Rearranging it to solve for y, we have y = Cxf(x) - x.
Finally, we can use the initial condition y(1) = 4 to find the value of C. Substituting x = 1 and y = 4 into the equation, we get 4 = Cf(1) - 1. Solving for C, we find C = (5 - f(1))/f(1).
Therefore, the explicit solution of the initial value problem is y = (5 - f(1))/f(1)x - x, where f(x) = ln|x| and C = (5 - f(1))/f(1).