Question

solver+s-6t=cos(2x+y)

Answer 1

Answer:

Solve differential equation y′′+y=cos2xy″+y=cos2⁡x

We are looking at a homogeneous equation:

y′′+y=0y″+y=0

λ2+1=0⇒λ1=−i,λ2=i⇒yh=C1e0xcosx+C2e0xsinx=C1cosx+C2sinxλ2+1=0⇒λ1=−i,λ2=i⇒yh=C1e0xcos⁡x+C2e0xsin⁡x=C1cos⁡x+C2sin⁡x

Now we find a particular solution:

y′′+y=cos2x=12+cos(2x)y″+y=cos2⁡x=12+cos⁡(2x)

y′′+y=12y″+y=12

yp1=1/2

Answer 2

Answer is $f_1(y-3x)+f_2(y+2x)+\frac{y.cos(2x+y)}{4}+4(sin(2x+y))$

Given, $r+s-6t=cos(2x+y)$

We can also denote it as $\frac{\partial^2 z}{\partial x^2}+\frac{\partial^2 z}{\partial x \partial y}-6\frac{\partial^2 z}{\partial y^2}=y cos(2x+y)$

The symbolic form will be $(D^2+DD'-6D'^2)Z=ycos(2x+y)$

Its AE  is:

$\Rightarrow m^2+m-6=0\\\Rightarrow (m+3)(m-2)=0\\\therefore m=-3,m=2$

$\therefore CF=f_1(y-3x)+f_2(y+2x)$

$PI=\frac{1}{(D+3D')(D-2D')} . y\ cos(2x+y)\\\\=\frac{1}{(D-2D')}[\frac{1}{(D+3D')} . y\ cos(2x+y)]...(1)$

Now, we know slope m=-3 and function=$f_1(D-mD')$

we also know

$y=c-mx\\\therefore y=c+3x ...(2)$

Using (2) in (1)

$PI=\frac{1}{(D-2D')}[\frac{1}{(D+3D')} . y\ cos(2x+y)]\\=\frac{1}{(D-2D')}\int{(c+3x).cos(2x+y)dx}\\\\=\frac{1}{(D-2D')}(c+3x)\frac{sin(2x+y)}{2}-3(-cos(2x+y))\\\\=\frac{1}{(D-2D')}[(c+3x)\frac{sin(2x+y)}{2}+3cos(2x+y)]...(3)$

Now, we know slope m=+2 and function=$f_1(D-mD')$

we also know

$y=c-mx\\\therefore y=c-2x ...(4)$

Using (4) in (3)

$PI=cos(2x+y))\\\\=\frac{1}{(D-2D')}[\frac{y.sin(2x+y)}{2}+3cos(2x+y)]\\=\int{[(c-2x)\frac{sin(2x+y)}{2}+3cos(2x+y)]dx}\\=(c-2x)(\frac{-cos(2x+y)}{4}+(-2)(\frac{-sin(2x+y)}{2})+3sin(2x+y)\\=(-y)(\frac{-cos(2x+y)}{4}+4(sin(2x+y))\\=\frac{y.cos(2x+y)}{4}+4(sin(2x+y))$

Hence,

$Z=CF+PI\\\Rightarrow f_1(y-3x)+f_2(y+2x)+\frac{y.cos(2x+y)}{4}+4(sin(2x+y))$

to learn more about differential equations from the given link

https://brainly.in/question/6640235

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