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Solve using variation of parameters. A) y'' + y = 3 \sec x - x^2 + 1 B) 2y'' + y = \tan x +...

Question:

Solve using variation of parameters.

A) {eq}y'' + y = 3 \sec x - x^2 + 1 {/eq}

B) {eq}2y'' + y = \tan x + e^{2x} - 2 {/eq}

Differential equations.

Method of variation of parameter for particular solution:


For a differential equation

{eq}\hspace{30mm} \displaystyle y'' + P(x) y' + Q(x) y = f(x) {/eq}

If complementary solution is given by

{eq}\hspace{30mm} \displaystyle y_{c} = C_{1} y_{1} + C_{2} y_{2} {/eq}

Then the particular solution is given by

{eq}\hspace{30mm} \displaystyle \boxed{ y_{p} = - y_{1} \int \frac{ y_{2} f(x) }{ W(x) } dx + y_{2} \int \frac{ y_{1} f(x) }{ W(x) } dx } {/eq}

where {eq}W(x) {/eq} is the wronskian of independent solution {eq}y_{1} {/eq} and {eq}y_{2} . {/eq}


Answer and Explanation:


(a).

{eq}\hspace{30mm} \displaystyle y'' + y = 3 \sec x - x^2 + 1 {/eq}

Characteristic equation of the above differential equation is

{eq}\hspace{30mm} \displaystyle{ m^2 + 1 = 0 \\ m^2 = -1 } {/eq}

Roots of this equation is {eq}m = \pm i {/eq} complex, therefore the complementary solution is

{eq}\hspace{30mm} \displaystyle y_{c} = C_{1} \cos x + C_{2} \sin x {/eq}

Therefore the two linearly independent solution of the given differential equation is given by

{eq}\hspace{30mm} \displaystyle y_{1} = \cos x \quad \text{and} \quad y_{2} = \sin x {/eq}

Now by using method of variation of parameter the particular solution of the given differential equation is given by

{eq}\hspace{30mm} \displaystyle y_{p} = - y_{1} \int \dfrac{ y_{2} ( 3 \sec x - x^2 + 1 ) }{W(x) } dx + y_{2} \int \dfrac{ y_{1} ( 3 \sec x - x^2 + 1 ) }{W(x) } dx {/eq}

where {eq}W(x) {/eq} is the wronskian of independent solution {eq}y_{1} {/eq} and {eq}y_{2} . {/eq}

{eq}\hspace{30mm} \displaystyle{ W(x) = \begin{vmatrix} \cos x & \sin x \\ - \sin x & \cos x \end{vmatrix} \\ W(x) = 1 } {/eq}

Now the Particular solution is

{eq}\hspace{30mm} \displaystyle{ y_{p} = - \cos x \int \dfrac{ \sin x \ ( 3 \sec x - x^2 + 1 ) }{ 1 } dx + \sin x \int \dfrac{ \cos x \ ( 3 \sec x - x^2 + 1 ) }{ 1 } dx \\ y_{p} = - \cos x \int ( 3 \tan x - x^2 \sin x + \sin x ) dx + \sin x \int ( 3 - x^2 \cos x + \cos x ) dx \\ y_{p} = - \cos x \Big[ (x^2 - 2) \cos x - 2x \sin x - \cos x - 3 \ln | \cos x | \Big] + \sin x \Big[ 3x - (x^2-3) \sin x - 2x \cos x \Big] \\ y_{p} = 3 - x^2 + 3x \sin x + 3 \cos x \ln | \cos x | \\ \boxed{ y_{p} = 3 - x^2 + 3x \sin x + 3 \cos x \ln | \cos x | } \\ } {/eq}

Now the general solution is given by

{eq}\hspace{30mm} \displaystyle{ y(x) = y_{c} + y_{p} \\ \boxed{ y(x) = C_{1} \cos x + C_{2} \sin x + 3 - x^2 + 3x \sin x + 3 \cos x \ln | \cos x | } } {/eq}



(b).

{eq}\hspace{30mm} \displaystyle y'' + y = \tan x + e^{2x} - 2 {/eq}

Characteristic equation of the above differential equation is

{eq}\hspace{30mm} \displaystyle{ m^2 + 1 = 0 \\ m^2 = -1 } {/eq}

Roots of this equation is {eq}m = \pm i {/eq} complex, therefore the complementary solution is

{eq}\hspace{30mm} \displaystyle y_{c} = C_{1} \cos x + C_{2} \sin x {/eq}

Therefore the two linearly independent solution of the given differential equation is given by

{eq}\hspace{30mm} \displaystyle y_{1} = \cos x \quad \text{and} \quad y_{2} = \sin x {/eq}

Now by using method of variation of parameter the particular solution of the given differential equation is given by

{eq}\hspace{30mm} \displaystyle y_{p} = - y_{1} \int \dfrac{ y_{2} ( \tan x + e^{2x} - 2 ) }{W(x) } dx + y_{2} \int \dfrac{ y_{1} ( \tan x + e^{2x} - 2 ) }{W(x) } dx {/eq}

where {eq}W(x) {/eq} is the wronskian of independent solution {eq}y_{1} {/eq} and {eq}y_{2} . {/eq}

{eq}\hspace{30mm} \displaystyle{ W(x) = \begin{vmatrix} \cos x & \sin x \\ - \sin x & \cos x \end{vmatrix} \\ W(x) = 1 } {/eq}

Now the Particular solution is

{eq}\hspace{30mm} \displaystyle{ y_{p} = - \cos x \int \dfrac{ \sin x \ ( \tan x + e^{2x} - 2) }{ 1 } dx + \sin x \int \dfrac{ \cos x \ ( \tan x + e^{2x} - 2 ) }{ 1 } dx \\ y_{p} = - \cos x \int ( \sin x \tan x + e^{2x} \sin x - 2 \sin x ) dx + \sin x \int ( \sin x + e^{2x} \cos x - 2 \cos x ) dx \\ y_{p} = - \cos x \bigg[ \bigg( \frac{2}{5} e^{2x} -1 \bigg) \sin x + \bigg( 2 - \frac{1}{5} e^{2x} \bigg) \cos x - \ln | \cos(x/2) - \sin(x/2) | + \ln| \cos(x/2) + \sin(x/2) | \bigg] + \sin x \bigg[ \frac{1}{5} e^{2x} \sin x + \frac{2}{5} e^{2x} \cos x - 2 \sin x - \cos x \bigg] \\ y_{p} = \frac{e^{2x} }{5} - 2 + \cos x \ln | \cos(x/2) - \sin(x/2) | - \cos x \ln| \cos(x/2) + \sin(x/2) | \\ \boxed{ y_{p} = \frac{e^{2x} }{5} - 2 + \cos x \ln | \cos(x/2) - \sin(x/2) | - \cos x \ln| \cos(x/2) + \sin(x/2) | } \\ } {/eq}

Now the general solution is given by

{eq}\hspace{30mm} \displaystyle{ y(x) = y_{c} + y_{p} \\ \boxed{ y(x) = C_{1} \cos x + C_{2} \sin x + \frac{e^{2x} }{5} - 2 + \cos x \ln | \cos(x/2) - \sin(x/2) | - \cos x \ln| \cos(x/2) + \sin(x/2) | } } {/eq}




Learn more about this topic:

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Constant of Variation: Definition & Example

from High School Algebra II: Tutoring Solution

Chapter 19 / Lesson 10
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