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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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For the differential equations, find the particular solution satisfying the given condition: \[ \bigg[ x \: sin^2 \bigg( \frac{x}{y} - y \bigg) \bigg] dx + xdy = 0; y \frac{\large \pi}{\large 4} when \: x = 1\]

$\begin{array}{1 1}\cot \large(\large\frac{y}{x}) = \log \left | ex \right | \\ \cot \large(\large\frac{x}{y}) = \log \left | ex \right | \\ \tan \large(\large\frac{y}{x})= \log \left | ex \right | \\\cos \large(\large\frac{y}{x}) = \log \left | ex \right | \end{array} $

Can you answer this question?
 
 

1 Answer

+1 vote

 

Solution:

$ cot (y/x) = log |ex| $

 


 

Tool Box:

·        A differential equation of the form $dy/dx = F(x,y)$ is said to be homogenous if $F(x,y)$  is a homogenous function of degree zero.

·        To solve equations of these types, we make substitution $y = vx$ differentiating with respect to $x$ we get, $dy/dx = v + xdv/dx$

·        If the homogenous differential function is in the form $dx/dy= F(x,y)$ where $F(x,y)$ is homogenous function of degree zero, then we make substitution $x = vy$, then on differentiating we get $dx/dy = v + ydv/dy.$

 

Step1

Let us find out whether the given problem is a homogenous solution with the degree zero or not.

Let us rewrite the given problem as $dy/dx = -[xsin^2(x/y) - y]/x$

i.e., $F(x,y) = -[xsin^2(x/y) -y]/x$

$F(kx.ky) = -[kxsin^2(kx/ky) - ky]/kx = - [xsin^2(x/y) - y)]/x = k^0.F(x,y)$

Hence it is a differential equation with degree 0.

 

Step 2

Hence we substitute $y = vx$ and $dy/dx = v + xdv/dx$

Therefore $v+xdv/dx = -[xsin^2v - vx]/x$.

 

Cancelling the common factor $x$ we get,

$ v + xdv/dx = -[sin^2v-v] = v - sin^2v$.

 

­On cancelling v we get $dv/dx = - sin^2v$, bringing $dv$ and $v$ on the LHS and $x$ and $dx$ on the RHS, we get

$dv/sin^2v = -dx/x$, but $1/sin^2x = cosec^2x$

 

Therefore $cosec^2v dv = - dx/x$

Integrating on both sides $-cotv = - log|x| + logC$

 

Now substituting for $v$ as $y/x$ we get $cot(y/x) = logcx$

 

Step3

Given $y = ¶/4$ and $x = 1$, now substituting for $x$ and $y$ we get $cot(¶/4) = log|Cx1|$, we know cot $¶/4 = 1$

 

Hence $1 = logc$ or $c = e^1$ or $c = e$

 

Now substituting for C we get the solution as $cot(y/x) = log|ex|$

answered Dec 17, 2012 by vijayalakshmi_ramakrishnans
edited Dec 25, 2012 by balaji.thirumalai
 

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