Question

find the general solution of the differential equation using undetermined coefficients, y^(iv) -2y"'+2y"=3e^-x+2e^-x.x+e^-sinx.

Answer 1

find the general solution of the differential equation using undetermined coefficients, y^(iv) -2y"'+2y"=3e^-x+2e^-x.x+e^-sinx.

Solution:

We can solve differential equation using undetermined coefficients if the right (heterogeneous) part of the equation is in the form:

$e^nP_m$ , where $P_m$ is a polynomial of degree $m$;

or in the form:

$e^\alpha(P_k\cos{(\beta x)}+P_l\sin{(\beta x)})$ , where $P_k$ and $P_l$ are polynomials of degrees $k$ and $l$.

Since $e^{-\sin{x}}$ cannot be presented as $e^nP_m$ or $e^\alpha(P_k\cos{(\beta x)}+P_l\sin{(\beta x)})$, we can conclude that a question consists mistake. Assume the correct equation is the following:

$y^{IV}-2y'''+2y''=3e^{-x}+2e^{-x}\cdot x+e^{-x}\cdot \sin{x}$ (*)

Let's find the complementary solution to this differential equation:

$y^{IV}-2y'''+2y''=0$

The characteristic equation for this differential equation is the following:

$r^4-2r^3+2r^2=0$

$r^2(r^2-2r+2)=0$

$r=0$ ; $r=1\pm i$

The complementary solution:

$y_c=C_1+C_2\cdot x+e^{x}(C_3\cos{x}+C_4\sin{x})$

Let's proceed with finding a particular solution. The form of the particular solution would be:

$y_p=e^{-x}(Ax+B+C\cos{x}+D\sin{x})$

$y_p'=e^{-x}$$(A-Ax-B+(D-C)\cos{x}-(C+D)\sin{x})$

$y_p''=e^{-x}$$(Ax-2A+B-2D\cos{x}+2C\sin{x})$

$y_p'''=e^{-x}$$(-Ax+3A-B+2(C+D)\cos{x}+2(D-C)\sin{x})$

$y_p^{IV}=e^{-x}$$(Ax-4A+B-4C\cos{x}-4D\sin{x})$

Substituting $y_p^{IV}$, $y_p'''$ and $y_p''$ into the (*) gives:

$e^{-x}(Ax-4A+B-4C\cos{x}-4D\sin{x})-$$2e^{-x}(-Ax+3A-B+2(C+D)\cos{x}$ $+2(D-C)\sin{x})+$ $2e^{-x}(Ax-2A+B-2D\cos{x}+2C\sin{x})=$ $3e^{-x}+2e^{-x}\cdot x+e^{-x}\cdot \sin{x}$

Choose $A$ , $B$ , $C$, and $D$ so that the coefficients of the exponentials on either side of the equal sign are the same:

$A=\frac25$ , $B=\frac{43}{25}$ , $C=\frac{1}{16}$, $D=-\frac{1}{16}$.

The particular solution:

$y_p=e^{-x}(\frac25x+\frac{43}{25}+\frac{1}{16}\cos{x}-\frac{1}{16}\sin{x})$

General solution:

$y=y_c+y_p=$ $C_1+C_2\cdot x+e^{x}(C_3\cos{x}+C_4\sin{x})+$ $e^{-x}(\frac25x+\frac{43}{25}+\frac{1}{16}\cos{x}-\frac{1}{16}\sin{x})$

Solution:

$y=C_1+C_2\cdot x+e^{x}(C_3\cos{x}+C_4\sin{x})+$ $e^{-x}(\frac25x+\frac{43}{25}+\frac{1}{16}\cos{x}-\frac{1}{16}\sin{x})$