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Solve the following. $(y - x)\large\frac{dy}{dx}$=$a^{2}$

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1 Answer

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Toolbox:
  • Linear Differential equation.
  • This is of the form $\large\frac{dy}{dx}$$+Py=Q$ where P and Q are functions of x only.
  • The integrating factor $I= e^{\int \large pfd}$ and the $G.S$ is$ ye^{\large pdx}=\int Q e^{\large pdx} dx+c$
Step 1:
$(y-x) \large\frac{dy}{dx}$$=a^2$
=>$a^2 \large\frac{dx}{dy}$$=y-x$
Step 2:
$a^2\large\frac{dx}{dy}$$+x=y$ divided by $a^2$
This is linear in $\large\frac{dx}{dy}$
$\large\frac{dx}{dy}+\frac{x}{a^2}=\frac{y}{a^2}$
$P(y)=\frac{1}{a^2}$$, 8 (y)=\frac{y}{a^2}$
$e^{\large \int pdy}=e^{\int \large\frac{1}{a^2} dy}$
$\qquad=e^{\Large\frac{y}{a^2}}$
Step 3:
The solution is $xe^{\large\frac{y}{a^2}}=\int \large\frac{y}{a^2} e^{\large\frac{y}{a^2}}$$dy+c$
$xe^{\large\frac{y}{a^2}}=\large \frac{1}{a^2}$$\bigg [y. a^2e^{\large\frac{y}{a^2}}-a^4e^{\large\frac{y}{a^2}}\bigg]+c$
$xe^{\large\frac{y}{a^2}}=y \;e^{\large\frac{y}{a^2}}-a^2e^{\large\frac{y}{a^2}}+c$
answered Sep 6, 2013 by meena.p
 
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