Answer to Question #126953 in Differential Equations for poulami paul

Question

(2x+1)(x+1)d^2y/dx^2+2xdy/dx-2y=(2x+1)^2

Answer 1

$(2x+1)(x+1)y''(x)+2xy'(x)-2y=(2x+1)^2$  

Ordinary differential equation classification: second-order linear ordinart differential equation.

Step by step solution:

1-step: Substitute $-2= \tfrac{d}{dx} (2x) - \tfrac{d^2}{dx}((x+1)(2x+1)):$

$\tfrac{d^2 y(x)}{dx^2} (x+1)(2x+1)+2x \tfrac{dy(x)}{dx}+( \tfrac{d}{dx}(2x)- \tfrac{d^2}{dx^2}((x+1)(2x+1)))y(x) = (2x +1)^2$

2-step: Expand

$\tfrac{d^2 y(x)}{dx^2}(x+1)(2x+1)+2x\tfrac{dy(x)}{dx}+\tfrac{d}{dx}(2x)y(x)-\tfrac{d^2}{dx^2}((x+1)(2x+1))y(x)=(2x+1)^2$

3-step: Add and subtract $\tfrac{dy(x)}{dx}\tfrac{d}{dx}((x+1)(2x+1))$ to the left hand sinde:

$\tfrac{dy(x)}{dx}\tfrac{d}{dx}((x+1)(2x+1))+\tfrac{d^2y(x)}{dx^2}(x+1)(2x+1)+2x\tfrac{dy(x)}{dx}+\tfrac{d}{dx}(2x)y(x)-\tfrac{dy(x)}{dx}\tfrac{d}{dx}((x+1)(2x+1))-\tfrac{d^2}{dx^2}((x+1)(2x+1))y(x)=(2x+1)^2$

4-step: Apply the reverse product rule $f\tfrac{dg}{dx}+g\tfrac{df}{dx}=\tfrac{d}{dx}(fg)$ to the left-hand side:

$\tfrac{d}{dx}(\tfrac{dy(x)}{dx}(x+1)(2x+1))+\tfrac{d}{dx}(2xy(x))-\tfrac{d}{dx}((2x+2(x+1)+1)y(x))=(2x+1)^2$

5-step: Factor:

$\tfrac{d}{dx}(\tfrac{dy(x)}{dx}(x+1)(2x+1)+2xy(x)-(2x+2(x+1)+1)y(x))=(2x+1)^2$

6-step: Integrate both sides with respect to x:

$\int\tfrac{d}{dx}(\tfrac{dy(x)}{dx}(x+1)(2x+1)+2xy(x)-(2x+2(x+1)+1)y(x))dx=\int(2x+1)^2dx$

7-step: Evalute the integrals:

$\tfrac{dy(x)}{dx}(x+1)(2x+1)+2xy(x)-(2x+2(x+1)+1)y(x)=\tfrac{4x^3}{3}+2x^3+x+$ C₁

where C₁ is an arbitrary constant.

8-step: Rewrite the equation:

$\tfrac{dy(x)}{dx}+\tfrac{-2(x+1)-1)y(x)}{(x+1)(2x+1)}=\tfrac{-\tfrac{4x^3}{3}-2x^3-x-C}{(x+1)(2x+1)}$

9-step: Let $\mu(x)=exp(\int\tfrac{-2(x+1)-1}{(x+1)(2x+1)}dx)= \tfrac{x+1}{(2x+1)^2}$

Multiply both sides by $\mu(x):$

$\tfrac{(x+1)\tfrac{dy(x)}{dx}}{(2x+1)^2}+\tfrac{(-2(x+1)-1)y(x)}{(2x+1)^3}=-\tfrac{-\tfrac{4x^3}{3}-2x^3-x-C}{(2x+1)^3}$

10-step: Substitute $\tfrac{-2(x+1)-1}{(2x+1)^3}=\tfrac{d}{dx}(\tfrac{x+1}{(2x+1)^2}):$

$\tfrac{(x+1)\tfrac{dy(x)}{dx}}{(2x+1)^2}+\tfrac{d}{dx}(\tfrac{x+1}{(2x+1)^2})y(x)=-\tfrac{-\tfrac{4x^3}{3}-2x^3-x-C}{(2x+1)^3}$

11-step: Apply the reverse product rule $f\tfrac{dg}{dx}+g\tfrac{df}{dx}=\tfrac{d}{dx}(fg)$ to the ledt-hand side:

$\tfrac{d}{dx}( \tfrac{(x+1)y(x)}{(2x+1)^2}= -\tfrac{-\tfrac{4x^3}{3}-2x^3-x-C}{(2x+1)^3}$

12-step: Integrate both sides with respect to x:

$\int\tfrac{d}{dx}(\tfrac{(x+1)y(x)}{(2x+1)^2})dx=\int-\tfrac{-\tfrac{4x^3}{3}-2x^3-x-C_{1}}{(2x+1)^3} dx$

13 - step: Evaluate the integrals:

$\tfrac{(x+1)y(x)}{(2x+1)^2}=\tfrac{1}{3}(\tfrac{1}{4}(2x+1)+\tfrac{-6C_{1}+1}{8(2x+1)^2})+$C₂

where C₂ is an arbitrary constant.

14-step: Divide both sides by $\mu(x)=\tfrac{x+1}{(2x+1)^2}:$

$y(x)=\tfrac{(2x+1)^2(\tfrac{1}{12}(2x+1)+\tfrac{-6C_{1}+1}{24(2x+1)^2}+C_{2})}{x+1}$

15-step: Simplify the arbitrary constants:

$y(x)=\tfrac{(2x+1)^2(\tfrac{1}{12}(2x+1)+\tfrac{C_{1}}{(2x+1)^2}+C_{2})}{x+1}$

Answer:
$y(x) = \tfrac{C_{2}\sqrt{\smash[b]{2x+1}}x}{\sqrt{\smash[b]{-2x-1}}}+\tfrac{C_{1}\sqrt{\smash[b]{2x+1}}x}{\sqrt{\smash[b]{-2x-1}}(x+1)}+\tfrac{2x^3}{3(x+1)}+\tfrac{x^2}{2(x+1)}$