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# Find the particular solution satisfying the given condition $x^2dy+(xy+y^2)\;dx=0;y=\;1;\;when\;x=\;1$

$\begin{array}{1 1} (A)\;y + 2x = 3x^2 \\(B)y - 2x = 3x^2 \\ (C)\;y + x = 3x^2 \\(D)\;y + 4x = 3x^2 \end{array}$

Toolbox:
• To solve homogenous differnetial equation put $y = vx$ and $\large\frac{dy}{dx}$$= v+x\large\frac{dv}{dx} Step 1: Let us rearrange and write the equation as x^2dy = - (xy+y^2)dx Or \large\frac{dy}{dx}=-\large\frac{(xy+y^2)}{x^2} Using the information in the tool box, v + x\large\frac{dv}{dx }= -\frac{ (x.vx+v^2x^2)}{x^2} Taking x^2 as the common factor and cancelling in the RHS we get v+x\large\frac{dv}{dx}$$=- (v+v^2)$
Bringing v from the LHS to RHS we get
$x\large\frac{dv}{dx }$$= -v(v+2) Seperating the variables we get, \large\frac{dv}{v(v+2) }= -\large\frac{ dx}{x } Step 2: Integrating on both sides we get, \int \large\frac{dv}{v(v+2)} = - \int\large\frac{dx}{x} Integration of \large\frac{dv}{v(v+2)} can be done by the method of partial fraction \large\frac{A}{v }+\frac{ B}{(v+2)} = \frac{1}{v(v+2)} A(v+2) + B(v) = 1 Equating the coefficients of like terms we get, A+B = 0(equating for v) 2A = 1 (equating for constant term) or A = 1/2 and B = -1/2 Substituting this for A and B we get, \int [\large\frac{1}{2(v)} -\frac{ 1}{2(v+2)}] = - \int\large\frac{ dx}{x} \bigg[\large\frac{1}{2}\bigg]$$\log v - \bigg[\large\frac{1}{2}\bigg]$$\log (v-2) = \log x + log C \large\frac{1}{2}$$\log \large\frac{v}{(v+2) }=$$\log Cx \log\large\frac{v}{(v+2)}$$ = 2\log Cx$
$\log\bigg[\large\frac{v}{(v+2)}\bigg] =$$\log c^2x^2 \large\frac{v}{(v+2)} = c^2x^2 Substituting for v =\large\frac{ y}{x} we get, \Large\frac{(\Large\frac{y}{x})}{[(\Large\frac{y}{x}) + 2] }=$$C^2x^2$
$\large\frac{x^2y}{(y+2x)}$$= C^2 Step 3: Given y = 1 and x = 1 substituting for x and y to evaluate the value of C we get \large\frac{1}{3 }=$$ C^2$
Substituting for $C^2$ we get,
$\large\frac{x^2}{(y+2x)} =\frac{ 1}{3}$
$y + 2x = 3x^2$
This is the required equation.
edited Aug 1, 2013