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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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Step-by-step explanation:

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Studentka2010 [4]

Answer:

4x^{3} y^{2} (\sqrt[3]{4 x y})

Step-by-step explanation:

Another complex expression, let's simplify it step by step...

We'll start by re-writing 256 as 4^4

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Then we'll extract the 4 from the cubic root.  We will then subtract 3 from the exponent (4) to get to a simple 4 inside, and a 4 outside.

\sqrt[3]{4^{4} x^{10} y^{7} } = 4 \sqrt[3]{4 x^{10} y^{7} }

Now, we have x^10, so if we divide the exponent by the root factor, we get 10/3 = 3 1/3, which means we will extract x^9 that will become x^3 outside and x will remain inside.

4 \sqrt[3]{4 x^{10} y^{7} } = 4x^{3} \sqrt[3]{4 x y^{7} }

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4x^{3} \sqrt[3]{4 x y^{7} } = 4x^{3} y^{2} \sqrt[3]{4 x y}

The answer is then:

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