Question

1. a) Solve the following linear ODE. dy * dx + 2y = 4x2, x > 0 b) Solve the following ODE using the substitution, u = dy (x - y) dx = y c) Solve the Bernoulli's ODE dy 1 + -y = dx = xy2 ; x > 0

Answer 1

Task: solve x y'(X) + 2 y(x) = 4x? Answer: C1 y(x) = +x2 Ordinary differential equation classification: first-order linear ordinary differential equation Step by step solution: Solve the linear equation x dy(x) + 2 y(x) = 4x2: dx Divide both sides by x: dy(x) 2 y(x) = 4x dx + X Let y(x) = @[2/x dx = x. Multiply both sides by p(x): dy(x) +(2 x) y(x) = 4 x3 dx x2 d Substitute 2 x = dx dy(x) d + et (?) y(x) = 4x dx dg Apply the reverse product rule f dx df + g d Of g) to the left-hand side: dx dx ya) = 4x dx Integrate both sides with respect to x: d y(x)) dx = S 62 - S 4x° ax Evaluate the integrals: x y(x) = x+ +C, where cy is an arbitrary constant. Divide both sides by u(x) = x²: Answer: x+ +C1 y(x) = x2

Q2.

dy(x) Solve (x - y(x)) dx = y(x): dv(x) =X + V(x): dy(x) Let y(x) = x V(X), which gives dx dv(x) (X-XV(x)) + V = XV(X) dx (x)) (x : dv(x) Solve for dx d V(x) V(x)2 dx X (V(X) - 1) V(X) Divide both sides by V(X) - 1 1 dv(x) (V(X) - 1) dx V(x)2 X Integrate both sides with respect to x: (V(X) - 1) 1 dx= dx V(x)2 dy(x) dx Evaluate the integrals: 1 log(v(x)) + = -log(x) + C1, where cy is an arbitrary constant. V(X) Solve for víx): 1 V(X) = - W(- 1 x) Simplify the arbitrary constants: V(x) = - WCX) Substitute back for y(x) = x V(X): Answer: y(x) = WC x)

now lets do Q3

y(x) solve y'(X) + = x y(x)2 х Answer: y(x) = X (C1 + x) Ordinary differential equation classification: first-order nonlinear ordinary differential equation Step by step solution: y(x) Solve Bernoulli's equation d y(x) dx x y(x): х Divide both sides by – y(x)?: dy(x) 1 = -X y(x)2 x y(x) dy(x) 1 yox)2 dv(x) Let V(x) = which gives y(x)' dx dv(x) V(X) =-X dx x 11- Let u(x) = J FS-1/x dx 1 Multiply both sides by p(x): dux) V(X) = -1 х x2 dx 1 d Substitute - dx) dx 60 Qur d V(x) = -1 х dx + g df dx = d of g) to the left-hand side: dx dg Apply the reverse product rule f dx d V(X) = -1 dx х Integrate both sides with respect to x: J7 Jax = -14x Evaluate the integrals: V(X) = -x +C1, where cy is an arbitrary constant. Divide both sides by p(x) = х V(X) = X(-X +C1) Solve for y(x): 1 1 y(x) = V(x) x2 --Cix Simplify the arbitrary constants: Answer: 1 y(x) = - x² +Cjx

Now for doubts ask in comments

i hope you understand

all questions done with mentioned methods

stay safe take care