Question

Solve the differential equation.
2ax+x^2dy/dx=a^2+2ax

Answer 1

$\textbf{Given:}$

$(2ax+x^2)\frac{dy}{dx}=a^2+2ax$

$\textbf{This D.E can be written as}$

$\frac{dy}{dx}=\frac{a^2+2ax}{2ax+x^2}$

$\frac{dy}{dx}=\frac{a^2+2ax}{x(x+2a)}$........(1)

$\text{First we resolve the R.H.S into partial fractions}$

$\frac{a^2+2ax}{x(x+2a)}=\frac{A}{x}+\frac{B}{x+2a}$

$\frac{a^2+2ax}{x(x+2a)}=\frac{A(x+2a)+Bx}{x(x+2a)}$

$\implies\;a^2+2ax=A(x+2a)+Bx$

$\text{Put x=0}$

$a^2=A(2a)$

$\implies\,A=\frac{a}{2}$

$\text{Put x=-2a}$

$-3a^2=B(-2a)$

$\implies\,B=\frac{3a}{2}$

$\therefore\frac{a^2+2ax}{x(x+2a)}=(\frac{a}{2})\frac{1}{x}+(\frac{3a}{2})\frac{1}{x+2a}$

$\text{Now (1) becomes}$

$\frac{dy}{dx}=(\frac{a}{2})\frac{1}{x}+(\frac{3a}{2})\frac{1}{x+2a}$

$dy=(\frac{a}{2})\frac{1}{x}+(\frac{3a}{2})\frac{1}{x+2a}\;dx$

$\text{Integrating on bothsides, we get}$

$\int\;dy=\frac{a}{2}\int\;\frac{1}{x}+\frac{3a}{2}\int\;\frac{1}{x+2a}\;dx$

$y=\frac{a}{2}log|x|+\frac{3a}{2}\;log|x+2a|+c$ $\text{ which is the required solution}$

Answer 2

Answer:

Step-by-step explanation:

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