Answer to Question #218834 in Calculus for Njabulo

Question

Solve the following differential equation subject to the given initial condition. (a)dy/d(theta)=ysin(theta); y(pi)=3. (b) x^2 dy/dx=y-xy; y(1)=1

Answer 1

(a) $\dfrac{dy}{d\theta}=y\sin\theta; y(\pi)=3$

Solution:

Separate the variables

$\dfrac{dy}{y}=\sin\theta d\theta$

$\displaystyle\int\dfrac{dy}{y}=\int\sin\theta d\theta+c$

$\displaystyle\ln|y|=-\cos\theta+c$

General solution is

$\displaystyle y=e^{-\cos\theta+c}$

Rewrite general solution as

$\displaystyle y=e^{-\cos\theta}\cdot e^{c}$

To find $e^c$ take initial conditions $y(\pi)=3$ and substitute $\theta=\pi; y=3$

$\displaystyle 3=e^{-\cos\pi}\cdot e^{c}$

$\displaystyle 3=e^{-(-1)}\cdot e^{c}$

$\displaystyle 3=e\cdot e^{c}$

$\displaystyle e^{c}=\dfrac{3}{e}$

$\displaystyle y=e^{-\cos\theta}\cdot\frac{3}{e}$

Particular solution is

$\displaystyle y=3e^{-\cos\theta-1}$

(b) $x^2 \dfrac{dy}{dx}=y-xy; y(1)=1$

Solution:

Separate the variables

$x^2 \dfrac{dy}{dx}=y(1-x)$

$\dfrac{dy}{y}=\dfrac{(1-x)}{x^2}dx$

$\displaystyle\int\frac{dy}{y}=\int\frac{(1-x)}{x^2}dx+c$

$\displaystyle\int\frac{dy}{y}=\int\left(\frac{1}{x^2}-\frac{x}{x^2}\right)dx+c$

$\displaystyle\int\frac{dy}{y}=\int\left(\frac{1}{x^2}-\frac{1}{x}\right)dx+c$

$\displaystyle\ln|y|=-\frac{1}{x}-\ln|x|+c$

$\displaystyle\ln|y|+\ln|x|=-\frac{1}{x}+c$

$\displaystyle\ln|yx|=-\frac{1}{x}+c$

$\displaystyle yx=e^{-\frac{1}{x}+c}$

General solution is

$\displaystyle y=\frac{e^{-\frac{1}{x}+c}}{x}$

Substitute $x=1, y=1$

$\displaystyle 1=\frac{e^{-\frac{1}{1}+c}}{1}$

$\displaystyle 1=e^{-1+c}$

$\displaystyle e^0=e^{-1+c}$

$\displaystyle 0=-1+c$

$c=1$

Particular solution is

$\displaystyle y=\frac{e^{-\frac{1}{x}+1}}{x}$