solving differential equation $\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$

Question

How would you solve this third order differential equation:

$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$

My first thought was to take a double integral:

$$\iint\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}dxdx=\iint{x^2+2x+2}dxdx$$ so: $$y+\frac{dy}{dx}=\frac{x^4}{12}+\frac{x^3}{3}+x^2+c_1x+c_2$$ This is what I got to but I am unsure beyond here

Answer 1

$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$ Substitute $z=y''$ $$z'+z=x^2+2x+2$$ $$(z-x^2-2)'+(z-x^2-2)=0$$ $$v'=-v$$ Where $v=z-x^2-2$

$$\implies \ln|v|=-x+K \implies v=K_1e^{-x} $$ $$\implies z=x^2+2+K_1e^{-x}$$ Integrate twice to get y $$\boxed{y=K_1e^{-x}+K_2x+K_3+x^2+\frac {x^4}{12}}$$

Answer 2

You are very near to the full solution. Now multiply both sides by $e^x$ and note that the left-hand side is $$e^x\left(y+\frac{dy}{dx}\right)=\frac{d}{dx}\left(e^xy(x)\right).$$ Can you take it from here?

Answer 3

Well, there are many ways you can you solve this DE. However, the simplest would be using the characteristic equation and since its a non-homogeneous DE, we need to find a particular solution as well. Finally we need to use the superposition principle to add the solution.

Now, in symbols and numbers $$ r^3 +r^2 = 0 \\ r = 0\; \text{or}\; r = -1 $$

Note that for $r =0$ it is a double root because every n-polynomial has n roots. Therefore, $$ y_c = c_1 + c_2x + c_3e^{-x} $$ now plugging $y_p = Ax^4 + Bx^3 + Cx^2$ in such a way that it satisfies the DE we get $$ 24Ax + B + 12Ax^2 + 6Bx +2C = x^2+2x+2 \\ 12Ax^2+ (24A + 6B)x +(B+2C) = x^2 +2x+2 $$

hence $A=\frac{1}{12}$, $B = 0$ and $C=1$ therefore $y_p = \frac{1}{12}x^4 + x^2$ and our general solution would be the sum, so $$ y = c_1+ +c_2x+ c_3e^{-x} +\frac{1}{12}x^4 +x^2 $$ Hope this helps !

Answer 4

Here is another simple way $$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$ Multiply both side by $e^x$ We get $$(y''e^x)'= (e^x(x^2+2))'$$ we can reduce the order by direct integration

$$y''e^x= e^x(x^2+2)+K_1$$ $$y"=x^2+2+K_1e^{-x}$$ Integrate $$y'=\frac {x^3}3+2x+K_1e^{-x}+K_2$$ Integrate again $$y=\frac {x^4}{12}+x^2+K_1e^{-x}+K_2x+K_3$$