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Tuesday, 5 August 2014

CHAPTER 12 - Examples on First Order Linear 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.
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