Question

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Answer 1

Question :-

Solve this differential equation ,

$\sf{x \red{ \frac{dy}{dx}} + \green{ {y}^{2}} = 9}$

Answer :-

$\sf{ \red{\frac{1}{2 \times 3} } log( \green{ \frac{3 + y}{3 - y}} ) = \orange{log(x)} + c} \: \:$

Explanation :-

Given equation is

$\rightarrow \: \sf{x \red{ \frac{dy}{dx}} + \green{ {y}^{2}} = 9} \\ \:$

we will solve this by variable separation method of differential equation,

means one side only one type of function ,

$\rightarrow \sf{x \: \red{ \frac{dy}{dx}} = 9 - \green{ {y}^{2} } }\\ \: \\ \\ \rightarrow \sf{\: \red{x }\:dy = \blue{ (9 - }\orange{{x}^{2}} )dx} \\ \\ \rightarrow \sf{ \green{ \frac{1}{9 - {y}^{2}} } = \red{ \frac{1}{x} dx}} \\ \\ \sf{ taking \int \: on \: both \: sides} \\ \\ \rightarrow \sf{ \green{ \int \: } \red{ \frac{1}{ {3}^{2} - {y}^{2} }} dy = \orange{ \int \: }\frac{1}{x} dx} \\ \\ \because \sf{\int \: \red{ \frac{1}{ {a}^{2} - {x}^{2} }} dx= \frac{1}{2a} \green{ log( \frac{a + x}{a - x} ) } + c} \\ \\ \therefore \\ \\ \rightarrow \sf{ \red{\frac{1}{2 \times 3} } log( \green{ \frac{3 + y}{3 - y}} ) = \orange{log(x)} + c} \\ \\ \rightarrow \sf{ \green{log( \frac{3 + y}{3 - y} ){}^{ \frac{1}{6} }} - log(x) = c \: \: ..(1)} \\ \\ \because \sf{ log(m) - log(n) = log( \frac{m}{n} ) } \\ \\ \rightarrow \: \sf{ log( \red{\frac{ \big( { \frac{3 + y}{3 - y} \big)}^{ \frac{1}{6} } }{x} }) = c} \\ \\ \sf{ hence }\: \\ \sf{ \blue{ put \: value \: of \: c \: in \: eq.(1)}}\\ \rightarrow \: \orange{ \sf{ log( \frac{3 + y}{3 - y} ){}^{ \frac{1}{6} } } = \green{ log(x) } \: +} \red{ log( \frac{ \big( { \frac{3 + y}{3 - y} \big)}^{ \frac{1}{6} } }{x} ) \: }$

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Answer 2

So we're given,

$x\dfrac {dy}{dx}+y^2=9$

So,

$x\dfrac {dy}{dx}=9-y^2\\\\\\x\cdot\dfrac {1}{dx}=\dfrac {9-y^2}{dy}\\\\\\\dfrac {dx}{x}=\dfrac {dy}{9-y^2}\\\\\\\dfrac {1}{x}\ dx=\dfrac {1}{9-y^2}\ dy$

Now, integrating both the sides of the equation,

$\displaystyle\int\dfrac {1}{x}\ dx=\int\dfrac {1}{9-y^2}\ dy\\\\\\\int\dfrac {1}{x}\ dx=\int\dfrac {1}{(3+y)(3-y)}\ dy\quad\longrightarrow\quad (1)$

Now, consider the RHS only.

$\displaystyle\int\dfrac {1}{(3+y)(3-y)}\ dy$

Let,

$\dfrac {1}{(3+y)(3-y)}=\dfrac {A}{3+y}+\dfrac{B}{3-y}\quad\longrightarrow\quad (2)$

for some constants $A$ and $B.$ Then,

$\dfrac {1}{(3+y)(3-y)}=\dfrac {A(3-y)+B(3+y)}{(3+y)(3-y)}$

Since the denominators are same, the numerators imply,

$A(3-y)+B(3+y)=1$

Taking $y=3,$ we get,

$A(0)+6B=1\\\\\\B=\dfrac {1}{6}$

Taking $y=-3,$ we get,

$6A+B(0)=1\\\\\\A=\dfrac {1}{6}$

Then (2) becomes,

$\dfrac {1}{(3+y)(3-y)}=\dfrac {1}{6(3+y)}+\dfrac{1}{6(3-y)}\\\\\\\dfrac {1}{(3+y)(3-y)}=\dfrac {1}{6}\left [\dfrac {1}{3+y}+\dfrac{1}{3-y}\right]$

Thus, in (1),

$\displaystyle\int\dfrac {1}{x}\ dx=\dfrac {1}{6}\int\left(\dfrac {1}{3+y}+\dfrac{1}{3-y}\right)\ dy\\\\\\\int\dfrac {1}{x}\ dx=\dfrac {1}{6}\int\dfrac {1}{3+y}\ dy-\dfrac {1}{6}\int\dfrac{-1}{3-y}\ dy\\\\\\\ln|x|+\ln(a)=\dfrac {1}{6}\ln|3+y|-\dfrac {1}{6}\ln|3-y|+\ln(b)\\\\\\\ln|ax|=\ln\left (b\left|\dfrac {3+y}{3-y}\right|^{\frac {1}{6}}\right)$

for some constants $a$ and $b.$

Taking antilogs on both sides,

$ax=b\left (\dfrac {3+y}{3-y}\right)^{\frac {1}{6}}\\\\\\b^6\left (\dfrac {3+y}{3-y}\right)=a^6x^6\\\\\\\dfrac {3+y}{3-y}=\dfrac {a^6}{b^6}x^6$

Replacing $\dfrac {a^6}{b^6}$ by $c,$

$\dfrac {3+y}{3-y}=cx^6\\\\\\\dfrac {y+3}{y-3}=\dfrac {cx^6}{-1}\quad\longrightarrow\quad (3)$

I think we may recall the componendo - dividendo rule.

$\dfrac {m}{n}=\dfrac {p}{q}\quad\iff\quad\dfrac {m+n}{m-n}=\dfrac {p+q}{p-q}$

Thus, (3) implies,

$\dfrac {y}{3}=\dfrac {cx^6-1}{cx^6+1}\\\\\\\boxed {\boxed {y=3\dfrac {(cx^6-1)}{(cx^6+1)}}}$

Hence Solved!

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