Each DE in Problems 15–22 is a Bernoulli equation.In Problems 15–20 solve the given differential equation by using anappropriate substitution.  Please explain and show all steps solve both  15) x dy/dx+y= 1/y^2 17) dy/dx=y(xy^3-1)

Find the value of the determinant D given by: a b 3 D = c d -4 6 9 9 where a = 2, b = 4, c = 9, d = 13 Give your answer as an integer.

answres are correct 100%, please tell me how to solve.

In Problem 3  use Euler’s method to obtain a four-decimalapproximation of the indicated value. First use h= 0.1 and then useh =0.05. Find an explicit solution for each initial-value problem andthen construct tables similar to Tables 2.6.3 and 2.6.4. 3) y'=y,  y(0)=1;    y(0)

Please. Answer parts please. Attempt Once. Use cosine and show in rads and degree.

Please. Answer parts please. Attempt Once. Use cosine and show in rad and degree (2 diff eq)

4. Solve the following differential equations: a- y" +zy' ty = x² b- y" -4y = Sinx <- 4"+zy! +y=ex ANS- 9- y = ex (4 x + c₂) +x²-4x+6 2X -2X b- y = Ge²x² + cze²x - is sinx -X C- y=Qe + cake + -X

Points for stress vs strain (in image) Assume the compressive concrete strength (f’c) is 3,000 lb/in2 (psi)Calculate a cubic function (3rd order polynomial – Ax3+Bx2+Cx+Constant)Use this function to create a function that describes the slope of the cubic function (the derivative of thecubic function). This new function allows you to calculate the tangent to any point along the curve. Thetangent is the modulus of elasticity (E).   The concrete code provides a formula to calculate E for concrete. That formula is:E = 57,000√??′, where f’c is in units of psi, and E is in units of psi.Use the derivative function you calculated to locate the point on the curve where the slope of the curvematches E using the concrete code formula. Express that stress point on the curve as a percentage ofthe compressive strength of the concrete. Now, calculate the secant modulus for the test case using 1,500 psi (50% f’c) as the arbitrary point onthe curve.Assume fracture occurs at the last point…

Use back-substitution to solve the system of linear equations. 2r+3y-3z=-4 - 8y – 7z=73 =-7 A (-8, -7, -3) B (-7, -8, -3) (5, -3, -7) (-3, -8, -7) E (-8,-3, -7)

s = 2  t = 0