Question

please solve it fast​

Answer 1

Consider,

$\longrightarrow y=c_1e^{ax}\cos(bx)+c_2e^{ax}\sin(bx)$

for some arbitrary constants $c_1$ and $c_2.$

Multiplying $\left(a^2+b^2\right)$ to both sides,

$\longrightarrow (a^2+b^2)y=(a^2+b^2)c_1e^{ax}\cos(bx)+(a^2+b^2)c_2e^{ax}\sin(bx)$

$\longrightarrow (a^2+b^2)y=(a^2c_1+b^2c_1)e^{ax}\cos(bx)+(a^2c_2+b^2c_2)e^{ax}\sin(bx)\quad\quad\dots(1)$

Differentiating $y$ wrt $x,$

$\longrightarrow y_1=\dfrac{d}{dx}\left[c_1e^{ax}\cos(bx)+c_2e^{ax}\sin(bx)\right]$

$\longrightarrow y_1=c_1\left[ae^{ax}\cos(bx)-be^{ax}\sin(bx)\right]+c_2\left[ae^{ax}\sin(bx)+be^{ax}\cos(bx)\right]$

$\longrightarrow y_1=\left(ac_1+bc_2\right)e^{ax}\cos(bx)+\left(ac_2-bc_1\right)e^{ax}\sin(bx)$

Multiplying 2a to both sides,

$\longrightarrow 2ay_1=\left(2a^2c_1+2abc_2\right)e^{ax}\cos(bx)+\left(2a^2c_2-2abc_1\right)e^{ax}\sin(bx)\quad\quad\dots(2)$

Differentiating $y_1$ wrt $x,$

$\longrightarrow y_2=\dfrac{d}{dx}\left[\left(ac_1+bc_2\right)e^{ax}\cos(bx)+\left(ac_2-bc_1\right)e^{ax}\sin(bx)\right]$

$\begin{aligned}\longrightarrow y_2&=\left(ac_1+bc_2\right)\left[ae^{ax}\cos(bx)-be^{ax}\sin(bx)\right]\\&+\left(ac_2-bc_1\right)\left[ae^{ax}\sin(bx)+be^{ax}\cos(bx)\right]\end{aligned}$

$\begin{aligned}\longrightarrow y_2&=\left(a^2c_1+2abc_2-b^2c_1\right)e^{ax}\cos(bx)\\&+\left(a^2c_2-2abc_1-b^2c_2\right)e^{ax}\sin(bx)\quad\quad\dots(3)\end{aligned}$

Without loss of generality, let,

• $A=e^{ax}\cos(bx)$

• $B=e^{ax}\sin(bx)$

Then (1) becomes,

$\longrightarrow (a^2+b^2)y=(a^2c_1+b^2c_1)A+(a^2c_2+b^2c_2)B\quad\quad\dots(1)'$

and (2) becomes,

$\longrightarrow 2ay_1=\left(2a^2c_1+2abc_2\right)A+\left(2a^2c_2-2abc_1\right)B\quad\quad\dots(2)'$

and (3) becomes,

$\longrightarrow y_2=\left(a^2c_1+2abc_2-b^2c_1\right)A+\left(a^2c_2-2abc_1-b^2c_2\right)B\quad\quad\dots(3)'$

Now, $(3)'-(2)'+(1)'$ gives,

$\begin{aligned}\longrightarrow\ \ &y_2-2ay_1+(a^2+b^2)y\\=\ \ &\left(a^2c_1+2abc_2-b^2c_1-2a^2c_1-2abc_2+a^2c_1+b^2c_1\right)A\\+\ \ &\left(a^2c_2-2abc_1-b^2c_2-2a^2c_2+2abc_1+a^2c_2+b^2c_2\right)B\end{aligned}$

$\longrightarrow\underline{\underline{y_2-2ay_1+(a^2+b^2)y=0}}$

Hence Verified!

Answer 2

Answer:

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