Question

Solve the equation r + s – 6t = ycosx
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Answer 1

Answer:

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Answer 2

Answer:

$z=\phi_{1}(y+2 x)+\phi_{2}(y-3 x)+\sin x-y \cos x$  is the required solution of the equation $r + s - 6t = ycosx$ .

Given:

The equation is given as  $r + s - 6t = ycosx$

Objective:

We have to solve the equation.

Step-by-step explanation:

Given differential equation in D, D' notations will be,

$F\left(D, D^{\prime}\right) z \equiv\left(D^{2}+D D^{\prime}-6 D^{\prime 2}\right) z=y \cos x \ldots$

Auxillary Equation is given by

$\mathrm{m}^{2}+\mathrm{m}-6=0$

Solve for m then write complementary function.

$\Rightarrow \quad(m+3)(m-2)=0$

$\Rightarrow \quad m=2,-3 .$

Complementary function is given by

$C.F.=\phi_{1}(y+2 x)+\phi_{2}(y-3 x)$

Particular integral is given by

$P.I.=\frac{1}{F\left(D, D^{\prime}\right)} y \cos x$

$=\frac{1}{D^{2}+D D^{\prime}-6 D^{\prime 2} }y \cos x$

$=\frac{1}{\left(D-2 D^{\prime}\right)\left(D+3 D^{\prime}\right)} y \cos x$

$=\frac{1}{5 D^{\prime}}\left(\frac{1}{D-2 D^{\prime}}-\frac{1}{D+3 D^{\prime}}\right) y \cos x$

$=\frac{1}{5} \frac{1}{D-2 D^{\prime}}\left(\frac{y^{2}}{2} \cos x\right)-\frac{1}{5\left(D+3 D^{\prime}\right)} \frac{y^{2}}{2} \cos x$

$=\frac{1}{10} \frac{1}{D}\left(1-\frac{2 D^{\prime}}{D}\right)^{-1}\left(y^{2} \cos x\right)-\frac{1}{10 D}\left(1+\frac{3 D^{\prime}}{D}\right)^{-1} y^{2} \cdot \cos x$

$=\frac{1}{10 D}\left(1+\frac{2 D^{\prime}}{D}+\frac{4 D^{\prime 2}}{D^{2}}-1+\frac{3 D^{\prime}}{D}-\frac{9 D^{\prime 2}}{D^{2}}\right) y^{2} \cos x$

$=\frac{1}{10 D}\left(\frac{5 D^{\prime}}{D}-\frac{5 D^{\prime 2}}{D^{2}}\right) y^{2} \cos x$

Differentiate accordingly then simplify.

$=\frac{1}{2}\left(\frac{1}{D^{2}} \cdot 2 y \cos x-\frac{1}{D^{3}} \cdot 2 \cos x\right)=\frac{1}{2} \cdot 2\left(y \frac{1}{D} \sin x-\frac{1}{D^{2}} \sin x\right)$

$=y(-\cos x)-\frac{1}{D}(-\cos x)$

$=-y \cos x+\sin x$

The required general solution of the differential equation $r + s - 6t = ycosx$  is given by

$z= C.F. + P.I.$

$z=\phi_{1}(y+2 x)+\phi_{2}(y-3 x)+\sin x-y \cos x$