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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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