Differential Equation $y'''-3y''=x^2-1$

Question

$y'''-3y''=x^2-1$ I'm not sure how to solve this. I tried to use the method of undetermined coefficients but the third derivative of $(x^2)-1$ is $0$. Can that be correct?

Answer 1

Put $y''=v$ then try to solve it.

Answer 2

As Saheb alluded to in his answer, that differential equation is nothing more than a first-order equation in $y''$, so you can treat $y''$ as a regular function, solve for it, and the find the second antiderivative of the answer you get by doing that; that will be $y$.

Answer 3

$$y'''-3y''=x^2-1 \implies R^3-3R^2=R^2(R-3)=0 \implies R=0,3$$ $$y_H=c_1e^{3x}+c_2+c_3x$$ For the particular solution you need a polynomial of degree at least 4

$$y_P=Bx^4+Cx^3+Dx^2$$

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Or just integrate directly $$y'''-3y''=x^2-1 $$ $$y''-3y'=\frac {x^3}3-x+K_1 $$ Integrate again $$y'-3y=\frac {x^4} {12}-\frac {x^2}2+K_1x+K_2 $$ Simple first ode now...

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Or consider it as first order if you substitute $y''=z$

$$y'''-3y''=x^2-1$$ $$z'-3z=x^2-1$$ $e^{-3x}$ as integrating factor $$(ze^{-3x})'=(x^2-1)e^{-3x}$$ Integrate $$ze^{-3x}=\int(x^2-1)e^{-3x}dx=\frac{e^{-3x}}3+\int x^2e^{-3x}dx$$ $$y''=\frac 13+e^{3x}\int x^2e^{-3x}dx$$ $$y''=\frac 13-\frac {x^2}{3}+\frac {2e^{3x}}3\int xe^{-3x}dx$$ $$y''=-\frac {x^2}{3}-\frac {2x}9+\frac 7{27}+K_1e^{3x}$$ Integrate twice to get the final solution $$y'=-\frac {x^3}{9}-\frac {x^2}{9}+\frac {7x}{27}++K_1e^{3x}+K_2$$ $$\boxed{y=-\frac {x^4}{36}-\frac {x^3}{27}+\frac {7x^2}{54}+K_1e^{3x}+K_2x+K_3}$$

Answer 4

According to undetermined coefficient, the characteristics equation is $$r^3-3r^2=0$$ $$r^2(r-3)-0$$ so the roots are $$x_0=0$$ $$x_1=0$$ $$x_2=3$$ the complementary solution is $$y_c=c_0+c_1x+c_2e^{3x}$$ the particular solution $$y_p=Ax^2+Bx+C\rightarrow Ax^4+Bx^3+Cx^2$$ then you can complete the solution