Please help with this equationQuestions not that Hard its just Im really tired

Mathematics

2 answers:

ella [17]2 years ago

6 0

Answer:

X =5y

Step-by-step explanation:

1 Subtract yy from both sides.

6x=4x+11y-y

6x=4x+11y−y

2 Simplify 4x+11y-y4x+11y−y to 4x+10y4x+10y.

6x=4x+10y

3 Subtract 4x4x from both sides.

6x-4x=10y

6x−4x=10y

4 Simplify 6x-4x6x−4x to 2x2x.

2x=10y

5 Divide both sides by 22.

x=\frac{10y}{2}

10y

6 Simplify \frac{10y}{2}

10y

to 5y5y.

x=5y

belka [17]2 years ago

6 0

$\tt 6x+y=4x+11y\\\\6x-4x=11y-y\\\\2x=10y\\\\y=0.2x$

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Burka [1]

Answer:

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Step-by-step explanation:

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5 0

3 years ago

4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25

jarptica [38.1K]

Answer with explanation:

$\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}$

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

$=e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}$

Multiplying both sides by Integrating Factor

$x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)$