Answer to Question #217976 in Differential Equations for Njabulo

Question

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

Answer 1

Solution.

1.

$$\frac{dy}{d\theta}=y\sin{\theta}, y(π)=3$$ $$\frac{dy}{y}=\sin{\theta}d\theta,$$ $$\int\frac{dy}{y}=\int\sin{\theta}d\theta,$$ $$\ln |y|=-\cos{\theta}+C,$$

where C is some constant.

If $y(π)=3,$ then $\ln 3=-\cos{π}+C.$

From here

$C=\ln 3-1.$

We will have

$\ln y=-\cos{\theta}+\ln 3-1,$ or$\ln y=\ln e^{-\cos{\theta}}+\ln 3-\ln e.$

Answer.

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

2.

$$x^2\frac{dy}{dx}=y-xy, y(1)=1$$ $$\frac{dy}{y}=\frac{1-x}{x^2}dx,$$ $$\int\frac{dy}{y}=\int\frac{1-x}{x^2}dx,$$ $$\int\frac{dy}{y}=\int\frac{1}{x^2}dx-\int\frac{dx}{x},$$ $$\ln y=-\frac{1}{x}-\ln x+C,$$

where C is some constant.

If $y(1)=1,$ then $\ln 1=-1-\ln1+C.$

From here C=1.

We will have

$\ln y=\ln e^{-\frac{1}{x}}-\ln x+\ln e,$ or$y=\frac{e^{1-\frac{1}{x}}}{x}.$

Answer.

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