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Tuesday, 5 August 2014

CHAPTER 13- Exact Differential Equations - Concept of Exact DEs

    Example: 1     

Solve the DE \dfrac{{dx}}{{dy}} + \dfrac{x}{y} = {y^2}
Solution: 1

Step-1

The I.F. is
I.F. = {e^{\int {P(y)dy} }}(y is the ‘independent’ variable in this DE)
 = {e^{\int {\dfrac{1}{y}dy} }}
 = {e^{\ln y}}
 = y

Step-2

Multiplying by the I.F. on both sides, we have
y\dfrac{{dx}}{{dy}} + x = {y^3}
 \Rightarrow \dfrac{d}{{dy}}(xy) = {y^3}
 \Rightarrow  d(xy) = {y^3}dy

Step-3

Integrating both sides gives
xy = \dfrac{{{y^4}}}{4} + C
     Example: 2     

Solve the DE \dfrac{{dy}}{{dx}} = {x^3}{y^3} - xy
Solution: 2

Step-1

We have,
\dfrac{{dy}}{{dx}} + xy = {x^3}{y^3}
Note that since the RHS contains the term y^3 this DE is not in the standard linear DE form. However, a little artifice can enable us to reduce this to the standard form.
Divide both sides of the equation by y^3:
\dfrac{1}{{{y^3}}}\dfrac{{dy}}{{dx}} + \dfrac{x}{{{y^2}}} = {x^3}\ldots (1)

Step-2

Substitute \dfrac{1}{y^2}=v:
 \Rightarrow  \dfrac{{ - 2}}{{{y^3}}}\dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{dx}}
 \Rightarrow  \dfrac{1}{{{y^3}}}\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{2}\dfrac{{dv}}{{dx}}\ldots (2)
Using (2) in (1), we have
\dfrac{{ - 1}}{2}\dfrac{{dv}}{{dx}} + xv = {x^3}
 \Rightarrow  \dfrac{{dv}}{{dx}} + ( - 2x)v =  - 2{x^3}\ldots (3)

Step-3

This is now in the standard first-order linear DE form. The I.F. is
I.F. = {e^{\int { - 2xdx} }} = {e^{ - {x^2}}}
Thus, the solutions to (3) is
v \times I.F. = \int {Q(x) \times (I.F)dx}
 \Rightarrow  v{e^{ - {x^2}}} =  - 2\int {{x^3}{e^{ - {x^2}}}dx}

Step-4

Performing the integration on the RHS by the substitution t=-x^2 and then using integration by parts, we obtain
v{e^{ - {x^2}}} = {e^{ - {x^2}}}({x^2} + 1) + C
 \Rightarrow \dfrac{1}{{{y^2}}}{e^{ - {x^2}}} = {e^{ - {x^2}}}({x^2} + 1) + C
This is the required general solution to the DE.

This example also tells us how to solve a DE of the general form:
\dfrac{{dy}}{{dx}} + P(x)y = Q(x){y^n}\ldots (4)
We divide by y^n on both sides :
\dfrac{1}{{{y^n}}}\dfrac{{dy}}{{dx}} + P(x){y^{ - n + 1}} = Q(x)
and then substitute {y^{ - n + 1}} = v and proceed as described in the solution above.
DEs that take the form in (4) are known as Bernoulli’s DEs.
—————————————————————————
In the last section, we discussed how the multiplication of the I.F. = {e^{\int {Pdx} }}on both sides of the linear DE
\dfrac{{dy}}{{dx}} + P(x)y = Q(x)
renders this into an exact DE. We now consider the general case of exact DEs. In particular, we want to see what condition must be satisfied in order that the DE
M(x,\,y)dx + N(x,\,y)dy = 0\ldots (1)
In order for this DE to be exact, it’s LHS must be expressible as the complete differential of some function f(x,y), i.e.
M(x,\,y)dx + N(x,\,y)dy = d\left( {f(x,\,y)} \right)\ldots (2)
Now since the function f(x,y) is a function of both x and y, its total differential is a sum of partical differentials with respect to x and y, i.e.,
df = \dfrac{{\partial f}}{{\partial x}}dx + \dfrac{{\partial f}}{{\partial y}}dy\ldots (3)
Comparing (1) and (2), we have
\dfrac{{\partial f}}{{\partial x}} = M,  \dfrac{{\partial f}}{{\partial y}} = N\ldots (4)
This gives
\dfrac{{{\partial ^2}f}}{{\partial y\partial x}} = \dfrac{{\partial M}}{{\partial y}},  \dfrac{{{\partial ^2}f}}{{\partial x\partial y}} = \dfrac{{\partial N}}{{\partial x}}
For continuous f(x,y),
\dfrac{{{\partial ^2}f}}{{\partial x\partial y}} = \dfrac{{{\partial ^2}f}}{{\partial y\partial x}}
and thus for the DE in (1) to be exact, we see that the necessary (and in fact sufficient) condition is
\dfrac{{\partial M}}{{\partial y}} = \dfrac{{\partial N}}{{\partial x}}
If this condition is satisfied, the DE in (1) reduces to
df(x,y) = 0
which upon integration leads to the required solution:
f(x,\,y) = C
As an example, consider the DE
3x(xy - 2)dx + ({x^3} + 2y)dy = 0\ldots (5)
We have,
M(x,\,y) = 3x(xy - 2)   \Rightarrow  \dfrac{{\partial M}}{{\partial y}} = 3{x^2}
N(x,\,y) = {x^3} + 2y   \Rightarrow  \dfrac{{\partial N}}{{\partial x}} = 3{x^2}
Since \dfrac{{\partial M}}{{\partial y}} = \dfrac{{\partial N}}{{\partial x}}, the DE is exact and hence we can find a function f(x,y)such that the DE is expressible as df(x,\,y) = 0. Let us try to explicitly find this function.
From (4), we have
\dfrac{{\partial f}}{{\partial y}} = N \Rightarrow  \dfrac{{\partial f}}{{\partial y}} = {x^3} + 2y
Integrating with respect to x, while treating y as a constant, we have
f(x,\,y) = {x^3}y - 3{x^2} + \phi (y)\ldots (6)
The function \phi (y) acts as the arbitrary constant of integration, since y is constant for this integration process.
From (4) again we have
\dfrac{{\partial f}}{{\partial y}} = N \Rightarrow   \dfrac{{\partial f}}{{\partial y}} = {x^3} + 2y
Evaluating \dfrac{{\partial f}}{{\partial y}} from (6), we have
{x^3} + \phi \prime(y) = \dfrac{{\partial f}}{{\partial y}} = {x^3} + 2y
 \Rightarrow  \phi \prime(y) = 2y
 \Rightarrow  \phi (y) = {y^2} + C\ldots (7)
Finally, substituting (7) into (6), we have
f(x,\,y) = {x^3}y - 3{x^2} + {y^2} + C
Thus, the solution to the DE in (5) is
f(x,\,y) = {\rm{constant}}
 \Rightarrow  {x^3}y - 3{x^2} + {y^2} = {\rm{constant}}\ldots (8)
You are urged to verify that (8) is indeed the required solution by differentiating(8) and observing that (5) is obtained.
     Example: 3     

Solve the DE 2xydx + ({x^2} + 3{y^2})dy = 0
Solution: 3

Step-1

First of all, notice that this DE is homogeneous :
\dfrac{{dy}}{{dx}} =  - \dfrac{{2xy}}{{{x^2} + 3{y^2}}}
The substitution y=vx leads to
v + x\dfrac{{dv}}{{dx}} = \dfrac{{ - 2v{x^2}}}{{{x^2} + 3{v^2}{x^2}}} = \dfrac{{ - 2v}}{{1 + 3{v^2}}}
 \Rightarrow  x\dfrac{{dv}}{{dx}} =  - v - \dfrac{{2v}}{{1 + 3{v^2}}}
 = \dfrac{{ - 3v - 3{v^3}}}{{1 + 3{v^2}}}
 = \dfrac{{ - 3v(1 + {v^2})}}{{1 + 3{v^2}}}
 \Rightarrow \dfrac{{1 + 3{v^2}}}{{v(1 + {v^2})}}dv =  - 3\dfrac{{dx}}{x}

Step-2

The substitution v^2=t leads to
\left( {\dfrac{{1 + 3t}}{{1 + t}} \cdot \dfrac{1}{{2t}}} \right)dt =  - 3\dfrac{{dx}}{x}
 \Rightarrow  \left( {\dfrac{1}{{1 + t}} + \dfrac{1}{{2t}}} \right)dt =  - 3\dfrac{{dx}}{x}

Step-3

Integrating both sides gives
\ln (1 + t) + \dfrac{1}{2}\ln t =  - 3\ln x + \ln C
 \Rightarrow  \sqrt t (1 + t){x^3} = C
 \Rightarrow  v(1 + {v^2}){x^3} = C
 \Rightarrow y({x^2} + {y^2}) = C
Solution: 3 (Alternate)

Step-1

We now solve this DE again using the exact differential approach since by observation this DE satisfies the required criterion for it to be exact. We have
\dfrac{{\partial f}}{{\partial x}} = M = 2xy, \dfrac{{\partial f}}{{\partial y}} = N = {x^2} + 3{y^2}
Integrating the first relation, we have
f(x,\,y) = {x^2}y + \phi (y)\ldots (1)

Step-2

Differentiating this w.r.t. y and comparing with the expression for \dfrac{{\partial f}}{{\partial y}} above, we have
\dfrac{{\partial f}}{{\partial y}} = {x^2} + \phi \prime(y) = {x^2} + 3{y^2}
 \Rightarrow  \phi \prime(y) = 3{y^2}
 \Rightarrow  \phi (y) = {y^3} + C'\ldots (2)
Thus, from (1) and (2),
f(x,\,y) = {x^2}y + {y^3} + C'

Step-3

The solution to the exact DE is
f(x,\,y) = {\rm{constant}}
 \Rightarrow  {x^2}y + {y^3} = C
 \Rightarrow  y({x^2} + {y^2}) = C
which is the same as the one obtained earlier. Thus, the exact differential approach might lead to the solution faster than the other approaches we’ve discussed earlier.
Sometimes, the fact that the DE is exact is evident merely be inspection. We list down such exact differentials (verify the truth of these relations):
xdy + ydx \Rightarrow  d(xy)
\dfrac{{xdy - ydx}}{{{x^2}}}  \Rightarrow  d\left( {\dfrac{y}{x}} \right)
\dfrac{{ydx - xdy}}{{{y^2}}} \Rightarrow    d\left( {\dfrac{x}{y}} \right)
\dfrac{{xdy - ydx}}{{{x^2} + {y^2}}} \Rightarrow d\left( {{{\tan }^{ - 1}}\dfrac{y}{x}} \right)
xdx + ydy \Rightarrow d\left( {\dfrac{{{x^2} + {y^2}}}{2}} \right)
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