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HW06: Problem 2 Previous Problem Problem List Next Problem (1 point) The equation бх? + Зу...

Question:

HW06: Problem 2 Previous Problem Problem List Next Problem (1 point) The equation бх? + Зу (*) ху can be written in the form

HW06: Problem 2 Previous Problem Problem List Next Problem (1 point) The equation бх? + Зу (*) ху can be written in the form y = f(y/x), i.e., it is homogeneous, so we can use the substitution u = equation with dependent variable u = u(x). yx to obtain a separable Introducing this substitution and using the fact that y xu' +u we can write (*) as = У = xu' u = f(u) where f(u) Separating variables we can write the equation in the form dx g(u) du = X where g(u) An implicit general solution with dependent variable u can be written in the form = C. x4+ y/x back into the variables x and y and using the initial condition y(1) = 1 we find Transforming u = C = Finally solve for y to obtain the explicit solution of the initial value problem y =

Answers

Sd! y 6x+Boy? - + ++ (3(say So (k) is homogeneous for forom ) => y = Ux of sy dy - y = u ex + 4.1 I So from c . y = xu tu - €ay glu) du = de where (say), . Now from (2), u du de 6 +242 => 4udu 4 dx 6 +242 - (6+24²). du 4 de (6 +24²) and (6+242) - 4 d2-4 ( 6+2 4 2) = C From the initial condition y (1) = 1 we get, ca 6 + 2.8 So we get 6274 + 2 y ² x 6 - 8

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