Solve the initial value problem y' = (x + y − 3)² with y(0) = 0. To solve this, we should use the substitution u = x+y 3 help (formulas) u' = 1+y' help (formulas) Enter derivatives using prime notation (e.g., you would enter y' for dy). After the substitution above, we obtain the following differential equation in x, u, u'. u'=1+u help (equations) The solution to the original initial value problem is described by the following equation in x, y. help (equations) Book: Section 1.5 of Notes on Diffy Qs

5. Find the derivative of y= x' +2x -1

(1 point) Solve the initial value problem yy' +x = x² + y with y(1) = -V3. a. To solve this, we should use the substitution U = x^2+y^2 help (formulas) u' = 2x+2yy' help (formulas) Enter derivatives using prime notation (e.g., you dy would enter y' for dx b. After the substitution from the previous part, we obtain the following linear differential equation in x, и, и'. 1/2u'=sqrtu help (equations) c. The solution to the original initial value problem is described by the following equation in x, y. y=+-sqrt(9x+9) help (equations)

1. Candace tried to find the derivative of the function y = h(x)e-4f(x) but her group members disagreed with her result. What did she get right? What did she get wrong? Correct her y' = h'(x)e-4f(x) + h(x)e-4f(x) (-4) work. After figuring out her error, Candace attempted to check her solution by rearranging the right side of the equation y = h(x)e-4f(x) into a fraction and using the quotient rule to differentiate. Verify the correct solution using this method.