User:Egm4313.s12.team13.Neisha.R2.1

Problem R 2.8B
'''pg.59 #15

Problem Statement:

15. Find a general solution. Check by substitution.

$$  \; {y}''+0.54 {y}'+(0.0729+ \pi )y=0 $$

Solution:

$$  \; {y}''+0.54 {y}'+(0.0729+ \pi )y= \frac{d^2y}{dx^2}\ + 0.54 \frac{dy}{dt}\ + (0.00729+ \pi )y=0   $$

If  $$ \; \frac{d}{dx}\ = \lambda $$,

$$ \; \lambda ^{2}+0.54 \lambda +(0.0729+ \pi)=0 $$

Solving for $$ \; \lambda $$ by using the quadratic formula,

$$ \lambda_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\,\!$$

$$ \lambda_{1} = -0.27 -{ \sqrt{ \pi } i} $$ $$ \lambda_{2} = -0.27 +{ \sqrt{ \pi } i} $$

In the standard form: $$ \; \lambda=a + bi $$.

General Solution:

The general homogeneous solution is:

$$ y= c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\ \!$$

When plugging in values for a and b, the general solution becomes:

$$ y= c_1 e^{-0.27 x} \cos \sqrt{ \pi }x + c_2 e^{-0.27 x} \sin \sqrt{ \pi }x\ \!$$

So $$ y=e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

Check by Substitution:

Solving for derivatives:

$$ y=e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

$$ \; y'=e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)-0.27e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

$$ \; y''=e^{-0.27 x}(- \pi c_1 \cos \sqrt{ \pi }x - \pi c_2 \sin \sqrt{ \pi }x)+ 0.0729e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)-0.54e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x) \ \!$$

Plugging in  $$   \; y, y', y $$ into $$   \; {y}+0.54 {y}'+(0.0729+ \pi )y=0 $$:

$$ \; [e^{-0.27 x}(- \pi c_1 \cos \sqrt{ \pi }x - \pi c_2 \sin \sqrt{ \pi }x)+ 0.0729e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)-0.54e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)]$$

$$ \; +0.54 [e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x +  \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)-0.27e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)]+(0.0729+ \pi )[e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.