Separable Ordinary Differential Equations

Solve the following differential equations or Initial Value Problems (IVP).

  1. \(\displaystyle{(x+1)\frac{dy}{dx}=xy}\) View Solution
  2. \(\displaystyle{\frac{dy}{dx}+y^2\sin (x) = 0}\) View Solution
  3. \(\displaystyle{\frac{dy}{dx}= \cos^2 (x)\, \cos^2 (2y)} \) View Solution
  4. \(\displaystyle{x \frac{dy}{dx} = (1-y^2)^{1/2}} \) View Solution
  5. \(\displaystyle{x\frac{dy}{dx}=\cos(\ln x) }\) View Solution
  6. \(\displaystyle{(x^2-9) \frac{dy}{dx} +x y = 0} \) View Solution
  7. \(\displaystyle{(y^2+1)\, dx = y \sec^2(x)\, dy} \) View Solution
  8. \(\displaystyle{\frac{dy}{dx} = y^2-4} \) View Solution
  9. \(\displaystyle{x^2\frac{dy}{dx}=\sec^2(1/x) }\) View Solution
  10. \(\displaystyle 2x^2\frac{dy}{dx}=\frac{\sin\left(\frac{1}{x}\right)}{e^{2y}}\) View Solution
  11. \(\displaystyle{x\frac{dy}{dx}=1+y^2, \: y(-1)=1 }\) View Solution
  12. \(\displaystyle{\frac{dy}{dx}- (\sin x) y = 2 \sin x, \: y(\pi/2)=1}\) View Solution
  13. \(\displaystyle{e^{-x^2}\frac{dy}{dx} = \frac{x}{y}, \: y(0)=-2}\) View Solution
  14. Find the solution of the given initial value problem (IVP) in explicit form and determine the interval in which the solution is defined. (a) \(\displaystyle{\frac{dy}{dx}=\frac{1-2x}{y}}, \: y(1)=-2\) View Solution (b) \(\displaystyle{\frac{dr}{d\theta}= r^2 /\theta, \: r(1)=2}\) View Solution (c) \(\displaystyle{\frac{dy}{dx} = xy^3(1+x^2)^{-1/2}, \: y(0)=1}\) View Solution
  15. Solve the ODE by making a suitable substitution. View Solution
    \(\displaystyle{\dfrac{dy}{dx}=(x+y)^2} \)