Physics, asked by Anonymous, 5 months ago

aNswer it ASAP ......!!! ​

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Answered by taniyath05
2

Answer: below is answer

Explanation:

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Answered by Anonymous
1

Explanation:

\star\:\:\bf\large\underline\blue{Question:-}

Question:−

Simplify the problem.

\star\:\:\bf\large\underline\blue{Solution:-}

Solution:−

\sqrt{a + b + 2x + 2 \sqrt{ab + (a + b)x + {x}^{2} } }

Now, let

y = \sqrt{ab + (a + b)x + {x}^{2} }y=

Therefore,

\sqrt{a + b + 2x + 2 \sqrt{ab + (a + b)x + {x}^{2} } }

= \sqrt{a + b + 2x + 2 + 2 \sqrt{y} }=

Now, we will factorise y.

\sqrt{ab + (a + b)x + {x}^{2} }

= \sqrt{ab + ax + bx + {x}^{2} }=

= \sqrt{a(b + x) +x(b+ x)}=

= \sqrt{(x + a)(x + b)}=

Now,

= \sqrt{a + b + 2x + 2 + 2 \sqrt{y} }=

= \small\sqrt{(x + a) + (x + b) + 2 \sqrt{(x + a)(x + b)} }=

= \sqrt{ {( \sqrt{x + a}) }^{2} + {( \sqrt{(x + b} )}^{2} + 2 \times \sqrt{(x + a} \times \sqrt{(x + b)} }=

= \sqrt{ {( \sqrt{x + a} + \sqrt{x + b} )}^{2} }=

= \sqrt{x + a} + \sqrt{x + b}=

\star\:\:\bf\large\underline\blue{Answer:-}

\sqrt{x + a } + \sqrt{x + b}✪QUESTION✪:-

Find the Solution of the given condition

\sf\dfrac{dy}{dx} - 3y \: cot \: x = sin2x,y = 2 \: when \: x = \dfrac{\pi}{2}

✪SOLUTION✪

•The given differential equation is a linear differential equation i.e, of the form \displaystyle\sf\dfrac{dy}{dx}+Py=Q

P=-3cot x and Q=sin 2 x

Here,⠀

\sf I.F.={ \sf e\:}^{ \displaystyle \sf\int \small P \: dx}={e}^{ \displaystyle \sf\int \small -3 \: cot \: x}

⠀⠀

\displaystyle \sf = {e}^{ - 3 \: log(sin \: x)} = {e}^{log(sin \: x) ^{ - 3} }=e

⠀⠀⠀⠀⠀⠀⠀⠀⠀\sf( Assuming\:0 < x < \pi)(Assuming0<x<π)

⠀⠀

\sf = {(sin \: x)}^{ - 3}=(sinx)

Hence,the solution of the given equation is given by

⠀⠀

\displaystyle \sf y{(sin \: x)}^{ - 3} = \int {(sin \: x)}^{ - 3} cos \: x \: dx+Cy(sinx)

= \displaystyle\sf2\int {(sin \: x)}^{ - 2} cos \: x \: dx=2∫(sinx)

\displaystyle \sf = \dfrac{2{(sin \: x)}^{ - 1} }{ - 1} + C=

⠀⠀\sf or \: y = - 2{(sin \: x)}^{3 - 1} + C \: {sin}^{3}xory=−2(sinx)

⠀⠀\sf or \: y = - 2{(sin \: x)}^{2} + C \: {sin}^{3}x .....(1)

When \sf x=\dfrac{\pi}{2},\:then\:y=2x=

⠀⠀

\sf \therefore2 = - 2{ sin}^{2} \bigg( \dfrac{ \pi}{2}\bigg) + C \:{sin}^{3} \bigg( \dfrac{\pi}{2} \bigg)

\implies \sf2 = - 2 + C⟹2=−2+C

⠀⠀

\implies \sf C=4⟹C=4</p><p></p><p>Hence the required solution is</p><p></p><p>[tex]\sf \: y = - 2{(sin \: x)}^{2} + 4{sin}^{3}x .......(from 1)

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