For a linear, inhomogenous ordinary differential equation:

$ \dfrac{dy}{dx} + 2y = 4x^2 +6x -1$

$(a)$ Find one solution to equation above that is quadratic. ie. of the form $P(x) = Ax^2 +Bx + C$

I used the integrating factor and find out that $y = 2x^2 +x -1$

$(b)$ solve the homogeneous linear equation $\dfrac{dy}{dx} + 2y = 0$

And my solution is $y = e^{-2x+C}$ or $y=0$

$(c)$ Solve $ \dfrac{dy}{dx} + 2y = 4x^2 +6x -1$ using $(a)$ and $(b)$. Hint: given a solution $y$ to $ \dfrac{dy}{dx} + 2y = 4x^2 +6x -1$, show that $y-P$ is a solution to a nice equation.

I don't really understand part $(c)$. Can anyone give me some hints?

  • $\begingroup$ The ODE is linear. If $y_k$, $k=1,2$, is such that $y_k'+2y_k=f_k$, then $y=y_1+y_2$ is such that $y'+2y=f_1+f_2$. $\endgroup$ – Pocho la pantera Sep 18 '13 at 21:36

If $L(y)$ is a linear differential operator, for example $L(y) = y'+2y$ as in your case, then every solution to $L(y) = f$ can be written as $$y = y_h + y_p$$ where $y_h$ is a solution to the homogeneous equation $L(y) = 0$ and $y_p$ is a particular solution to $L(y) = f$.

Proof. Let $y$ be an arbitrary solution to $L(y) = f$. Then $L(y-y_p) = L(y) - L(y_p) = f - f = 0$. Hence $y-y_p$ solves the homogeneous equation.

Conversely, if $y_h$ and $y_p$ are as above, then $y = y_h + y_p$ solves $L(y) = f$.

Proof. $L(y_h + y_p) = L(y_h) + L(y_p) = 0 + f = f$.

Notice the importance of $L$ being linear.


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