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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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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} $

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  • 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.
answered Aug 1, 2013 by sreemathi.v
edited Aug 1, 2013 by sreemathi.v

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