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2 years ago

Please help solve this equation?

Mathematics

1 answer:

finlep [7]2 years ago

5 0

${ \red{ \bold{cos \: y \: }}}$

Step-by-step explanation:

${ \green{ \tt{ \frac{1 \: + \: \cos \: y \: }{1 \: + \: \sec \: y \: }}}} \: → {eq}^{n} (1)$

But, as you know that

${ \blue{ \tt{sec \: y \:}}} = { \green{ \tt{\frac{1}{ \ \cos \: y }}}}$

Then the equation (1) becomes

${ \green{ \tt{ \frac{1 \: + \: cos \: y }{1 \: + \: \frac{1}{cos \: y} }}}} \:$

Multiply Numerator and Denominator by $\frac{cos \: y}{cos \: y}$

then,

${ \green{ \tt{( \frac{cos \: y}{cos \: y})}}} \: { \green{ \tt{ \frac{1 \: + \: cos \: y}{1 \: + \: \frac{1}{cos \: y}}}}}$

$= { \green{ \tt{ \frac{cos \: y \: + \: {cos}^{2} \: y }{cos \: y \: + \: 1 }}}}$

take cos y as common, then

${ \green{ \tt{cos \: y}}} \: { \green{ \tt( \frac{1 \: + \: cos \: y}{cos \: y \: + \: 1} )}}$

Here, (1+cos y/cos y + 1) gets cancelled.

Then the remaining answer is cos y.

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Step-by-step explanation:

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3 years ago

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$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

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Mashutka [201]

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