Question

Solve each differential equation by variation of parameters.

$\mathbf{3}{\mathbf{y}}^{\text{'}\text{'}}\mathbf{-}\mathbf{6}{\mathbf{y}}^{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\mathbf{secx}$.

Short answer

he required general solution of the given differential equation is $$\begin{gathered} c_1{e}^x\cos x+c_2{e}^x\sin x+ \\ \frac{1}{3}{e}^x\cos x\ln \left(\cos x\right)+\frac{1}{3}x{e}^x\sin x \end{gathered}$$.

Step-by-step solution

Note the given data

We need to find the general solution of the given differential equation.

We have second order equation $3y^{\text{'}\text{'}}-6y^{\text{'}}+6y=e^{x}secx$.

Finding the solution of the homogeneous solution

Finding the solution of the homogeneous equation

$y^{\text{'}\text{'}}-2y^{\text{'}}+2y=0$…………………………………………………………..(1)

Let $y=e^{mx}$be a solution of the given homogeneous solution

Differentiate the above equation with respect to x, we get

$y^{\text{'}}=m{e}^{mx}$………………………………………………………………(2)

Again, differentiate the above equation with respect to , we get

$y^{\text{'}\text{'}}=m^2{e}^{mx}$…………………………………………………………….(3)

Substitute the value$y=e^{mx}$ and equation (3) in differential equation (1), we get

$$\begin{gathered} m^2{e}^{mx}-2m{e}^{mx}+2e^{mx}=0 \\ {e}^{mx}\left(m^2-2m+2\right)=0 \\ {m}_{1,2}=\frac{2\pm \sqrt{4-4\times 1\times 2}}{2} \\ =\frac{2\pm \sqrt{-4}}{2} \end{gathered}$$

Then the roots are

$=1\pm i$

which are conjugate and complex.

Then the solution of a homogeneous solution is

$y_h=c_1{e}^{(1+i)x}+c_2{e}^{(1-i)x}$

$=\left(c_1+c_2\right)e^x\cos x+\left(c_1-c_2\right)e^x\sin x$

$=k_1{e}^x\cos x+k_2{e}^x\sin x$………………………………………………………………..(4)

Finding the Wronskian from left side

$$\begin{gathered} W\left(y_1,y_2\right)=\left|Y_1{Y}_2 \\ Y{\text{'}}_{2}Y{\text{'}}_2\right| \\ =\left|e^x\cos x{e}^x\sin x \\ {e}^x(\cos x-\sin x)e^x(\cos x+\sin x)\right| \\ =\left[e^x\cos x\right]\times \left[e^x(\cos x+\sin x)\right]-\left[e^x\sin x\right]\times \left[e^x(\cos x-\sin x)\right] \\ =e^{2x}{\cos }^{2}x+e^{2x}\sin x\cos x-e^{2x}\mathrm{sincos}x+e^{2x}{\sin }^{2}x \\ =\left(e^{2x}\left({\cos }^{2}x+{\sin }^{2}x\right)\right. \\ =e^{2x}\times \left(1\right) \\ =e^{2x} \end{gathered}$$

Finding the Wronskian from right side of given differential question

$$\begin{gathered} W_1=\left|0Y_2 \\ f\left(x\right)Y{\text{'}}_2\right| \\ =\left|0e^x\sin x \\ \frac{e^x}{3}{e}^x(\cos x+\sin x)\right| \\ =-\frac{1}{3}{e}^{2x}\sin x\times \frac{1}{\cos x} \\ =-\frac{1}{3}{e}^{2x}\tan x \\ {W}_2=\left|Y_{1}0 \\ Y{\text{'}}_{1}f\left(x\right)\right| \\ =\left|e^x\cos x0 \\ {e}^x(\cos x-\sin x)\frac{e^x\sec x}{3}\right| \\ =\frac{1}{3}{e}^{2x}\cos x\times \frac{1}{\cos x} \\ =\frac{1}{3}{e}^{2x} \end{gathered}$$

Finding the value of u1 and u2

$$\begin{gathered} u_1^{\text{'}}=\frac{W_1}{W\left(y_1,y_2\right)} \\ =\frac{-\frac{1}{3}{e}^{2x}\tan x}{e^{2x}} \\ =-\frac{1}{3}\tan x \\ {u}_2^{\text{'}}=\frac{W_2}{W\left(y_1,y_2\right)} \\ =\frac{\frac{1}{3}{e}^{2x}}{e^{2x}} \\ =\frac{1}{3} \end{gathered}$$

Integration

After that, we have to do integration for both u'₁and u'₂neglecting integration constants to obtain u₁ and u₂ as the following

$$\begin{gathered} u_1=\int u_1^{\text{'}}dx \\ =\int -\frac{1}{3}\tan xdx \\ =\frac{1}{3}\ln \left(\cos x\right) \\ {u}_2=\int u_2^{\text{'}}dx \\ =\int \frac{1}{3}dx \\ =\frac{1}{3}x \end{gathered}$$

Finding the required solution and we get

Substitute all the values of $u_1\text{and}{u}_2\text{and the functions}{y}_1\text{and}{y}_2$, we get the particular solution

$$\begin{gathered} y=y_h+y_p \\ =\left(\frac{1}{3}\ln \left(\cos x\right)\right)\times \left(e^x\cos x\right)+\left(\frac{1}{3}x\right)\times \left(e^x\sin x\right) \\ =\frac{1}{3}{e}^x\cos x\ln \left(\cos x\right)+\frac{1}{3}x{e}^x\sin x \end{gathered}$$

Finally, from equations (a) and (b) we can obtain the general solution of the differential equation $y^{\text{'}\text{'}}-2y^{\text{'}}+2y=\frac{e^x\sec x}{3}$as

$y=y_h+y_p$

Now, we find the general solution

$\sin x+\frac{1}{3}{e}^x\cos x\ln \left(\cos x\right)+\frac{1}{3}x{e}^x\sin x$