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labwork [276]
2 years ago
15

Find a particular solution to y" - y + y = 2 sin(3x)

Mathematics
1 answer:
leonid [27]2 years ago
8 0

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

     =e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}

Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

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Hey can anyone pls answer these math problems!!
Zepler [3.9K]

Answer:

1={5 and 3} 2={0} 3={0.209 and 4.791} 4={-3 and 4} 5={1 and 9} 6={-4.303 and -0.697}

Step-by-step explanation:

Your probably confused on what a, b, and c means.

The problems are written like this.

ax^2+bx+c=0

That formula finds the zeros.

For example 1:

x^2-8x+15=0\\ax^2-bx-c\\a=1\\b=-8\\c=15

Then use the formula

x=\frac{-b+-\sqrt{b^2-4ac} }{2a} \\\\x=\frac{-(-8)+-\sqrt{-8^2-4(1)(15)} }{2} \\x=\frac{8+-\sqrt{64-60} }{2} \\\\x=\frac{8+-2}{2}\\

If plus then 5, if minus then 3

CHECK:

x^2-8x+15=0\\5^2-8(5)+15=0\\25-40+15=0\\0=0\\3^2-8(3)+15=0\\9-24+15=0\\0=0

Hope this helps you understand how to do this by yourself.  Now to do the rest ;-;

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2 years ago
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