User:Egm6321.f11.team2.Xia/RP2.15

=Problem 2.15- Commutative partial derivations=

Given
Since $$\displaystyle P(x)y'+Q(x)y=R(x)$$ (2.15.1)

$$\displaystyle y'+\frac {Q(x)}{P(x)}y=\frac{R(x)}{P(x)}$$ (2.15.2)

is an L1-ODE_VC, there should be only one integration constant, not two.

Find
Show that the integration constant $$\displaystyle k_{1} $$ in $$\displaystyle h(x)=exp[\int^{x}a_{0}(s)ds+k_{1}]  $$ (2.15.3)

is not necessary, i.e., only $$ \displaystyle k_{2} $$ in $$ \displaystyle y(x)=\frac{1}{h(x)}[\int^{x}h(s)b(s)ds+k_{2}] $$ (2.15.4)

is necessary.

Solution
Substituting the eqn (2.15.3) to eqn (2.15.4), we obtain the following expression $$\displaystyle y(x)=\frac{1}{e^{k_{1}}\cdot e^{\int^{x}a_{0}(s)ds}}[e^{k_{1}}\cdot\int^{x}e^{\int^{x}a_{0}(s)ds}b(s)ds+k_{2}]$$ $$\displaystyle y(x)=\frac{1}{e^{\int^{x}a_{0}(s)ds}}[\int^{x}e^{\int^{x}a_{0}(s)ds}b(s)ds+\frac{k_{2}}{e^{k_{1}}}]$$ Then, we can define a new constant $$\displaystyle \bar{k}_{2} $$ such that $$\displaystyle \bar{k}_{2}=\frac{k_{2}}{e^{k_{1}}} $$ hence, this new constant contains $$\displaystyle k_{1} $$. In other words, $$\displaystyle k_{1} $$ is unnecessary.