Question

Identify the type of equations and Solve

a) 𝑦′ = 𝑥𝑦 1+𝑥2

b) sin 𝑦 𝑑𝑥 + cos 𝑦 𝑑𝑦 = 0

c) 𝑑𝑦 = 2𝑡 (𝑦2 + 9 )𝑑𝑡

Answer 1

$a)\newline 𝑦′ = \frac{𝑥𝑦}{1+𝑥^2}\newline \text{This linear ordinary Differential equation.}\newline\text{ Solving the given differential equation by variable separable form,}\newline \frac{dy}{y}= \frac{𝑥}{1+𝑥^2}dx\newline \text{ Integrate both side,}\newline \int\frac{dy}{y}= \frac{1}{2}\int\frac{2𝑥}{1+𝑥^2}dx\newline lny= \frac{1}{2}ln(1+x^2)+c\newline b)\newline sin 𝑦 𝑑𝑥 + cos 𝑦 𝑑𝑦 = 0\newline\text{ This linear ordinary Differential equation}.\newline\text{ Solving the given differential equation by variable separable form,}\newline \frac{cosy}{siny}dy= -dx\newline\text{ Integrate both side,}\newline \int\frac{cosy}{siny}dy= -\int dx\newline lnsiny=-x+c \newline c)\newline 𝑑𝑦 = 2𝑡 (𝑦^2 + 9 )𝑑𝑡\newline\text{ This linear ordinary Differential equation}.\newline\text{ Solving the given differential equation by variable separable form,}\newline 𝑑𝑦 = 2𝑡 (𝑦^2 + 9 )𝑑𝑡\newline \frac{1}{3^2+y^2}dy= 2tdt\newline \frac{1}{9} \frac{1}{1+(\frac{y}{3})^2}dy= 2tdt\newline\text{ Integrate both side,}\newline \frac{1}{9}\int \frac{1}{1+(\frac{y}{3})^2}dy= 2\int tdt\newline \frac{1}{3} tan^{-1}(\frac{y}{3})= t^2+c\newline$