Question

dy/dx-2y=cos3x. solve ................... quickly​

Answer 1

Answer:ye^-2t=e^-2t/13(3sin3x-2cos3x)+c

Step-by-step explanation:Please refer the following attachment for solution.

Answer 2

$ye^-^2^x=\frac{e^-^2^x}{13}(3sin3x-2cos3x)+C$

Step-by-step explanation:

Given,

$\frac{dy}{dx}-2y=cos3x$......(1)

Since this is a linear equation of the form

$\frac{dy}{dx}+Py=Q$.........(2)

On comparing equation (1) and (2)

P = -2,Q= cos3x

⇒$I.F = e^\int ^P^d^x = e^-^2^d^x=e^-^2^x$

On Multiplying both sides of equation(1),we get

$e^-^2^x.\frac{dy}{dx}-2ye^-^2^x=cos3xe^-^2^x$

By applying integration on both sides w.r.t x

$ye^-^2^x = \int e^-^2^xcos3xdx+C$

⇒$ye^-^2^x=I+C,where I=e^-^2^xcos3x.$.......(3)

Now,$I = \int e^-^2^xcos3xdx$

=$\frac{1}{3}e^-^2^xsin3x-\int \frac{(-2)}{3}e^-^2^xsin3xdx$

=$\frac{1}{3}e^-^2^xsin3x+\frac{2}{3}[\frac{-1}{3}e^-^2^xcos3x-\int (-2)e^-^2^x(\frac{-cos3x}{3})dx]$

$=\frac{1}{3}e^-^2^xsin3x+\frac{2}{3}[\frac{-1}{3}e^-^2^xcos3x-\frac{2}{3}\int e^-^2^xcos3xdx]$

=$\frac{1}{3}e^-^2^xsin3x-\frac{2}{9}e^-^2^xcos3x-\frac{4}{9}I$

⇒$(I+\frac{4}{9}I)=\frac{e^-^2^x}{9}(3sin3x-2cos3x)$

or$I=\frac{e^-^2^x}{13}(3sin3x-2cos3x)$

By putting the value of I inequation (3)

$\bold{ye^-^2^x=\frac{e^-^2^x}{13}(3sin3x-2cos3x)+C}$