Separable Ordinary Differential Equations:

Solve the following differential equations or IVP.

  1. \( \displaystyle{(x+1)\frac{dy}{dx}=xy}\)   solution

  2. \( \displaystyle{\frac{dy}{dx}+y^2\sin (x) = 0}\)   solution

  3. \( \displaystyle{\frac{dy}{dx}= \cos^2 (x)\, \cos^2 (2y)} \)   solution

  4. \(\displaystyle{x \frac{dy}{dx} = (1-y^2)^{1/2}} \)   solution

  5. \( \displaystyle{x\frac{dy}{dx}=\cos(\ln x) }\)  solution

  6. \( \displaystyle{(x^2-9) \frac{dy}{dx} +x y = 0} \)   solution

  7. \( \displaystyle{(y^2+1)  dx = y \sec^2(x)  dy} \)   solution

  8. \( \displaystyle{\frac{dy}{dx} = y^2-4} \) solution

  9. \(\displaystyle{x^2\frac{dy}{dx}=\sec^2(1/x) }\)   solution

  10. \(\displaystyle 2x^2\frac{dy}{dx}=\frac{\sin\left(\frac{1}{x}\right)}{e^{2y}}\) solution

  11. \(\displaystyle{x\frac{dy}{dx}=1+y^2, \: y(-1)=1 }\)   solution

  12. \( \displaystyle{\frac{dy}{dx}- (\sin x) y = 2 \sin x, \: y(\pi/2)=1}\)  solution

  13. \(\displaystyle{e^{-x^2}\frac{dy}{dx} = \frac{x}{y}, \: y(0)=-2}\)    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.

    15. (a) \( \displaystyle{\frac{dy}{dx}=\frac{1-2x}{y}}, \: y(1)=-2\)   solution

    (b) \(\displaystyle{\frac{dr}{d\theta}= r^2 /\theta, \: r(1)=2}\)   solution

    (c) \(\displaystyle{\frac{dy}{dx} = xy^3(1+x^2)^{-1/2}, \: y(0)=1}\)   solution


  15. Solve the ODE by making a suitable substitution. solution

    16. \(\displaystyle{\dfrac{dy}{dx}=(x+y)^2} \)