in problems 1 - 22 solve the given differential equation by separation of variables.\n1. $\frac{dy}{dx}=sin3x…

in problems 1 - 22 solve the given differential equation by separation of variables.\n1. $\frac{dy}{dx}=sin3x$\n2. $\frac{dy}{dx}=(x + 1)^2$\n3. $dx+e^{3x}dy = 0$\n4. $dy-(y - 1)^2dx = 0$\n5. $x\frac{dy}{dx}=4y$\n6. $\frac{dy}{dx}+2xy^2 = 0$\n7. $\frac{dy}{dx}=e^{3x + 2y}$\n8. $e^{x}y\frac{dy}{dx}=e^{-y}+e^{-2x - y}$\n9. $yln x\frac{dx}{dy}=(\frac{y + 1}{x})^2$\n10. $\frac{dy}{dx}=(\frac{2y + 3}{4x + 5})^2$\n11. $cos ydx+sec^{2}xdy = 0$\n12. $sin3xdx+2ycos^{3}3xdy = 0$\n13. $(e^{x}+1)^2e^{-y}dx+(e^{x}+1)^3e^{-x}dy = 0$\n14. $x(1 + y^{2})^{1/2}dx=y(1 + x^{2})^{1/2}dy$\n15. $\frac{ds}{dr}=ks$\n16. $\frac{dq}{dt}=k(q - 70)$\n17. $\frac{dp}{dt}=p - p^{2}$\n18. $\frac{dn}{dt}+n=nte^{t + 2}$\n19. $\frac{dy}{dx}=\frac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$\n20. $\frac{dy}{dx}=\frac{xy + 2y - x - 2}{xy - 3y + x - 3}$\n21. $\frac{dy}{dx}=xsqrt{1 - y^{2}}$\n22. $(e^{x}+e^{-x})\frac{dy}{dx}=y^{2}$\nin problems 23 - 30 find an explicit solution of the given initial - value problem.\n23. $\frac{dx}{dt}=4(x^{2}+1),x(pi/4)=1$\n24. $\frac{dy}{dx}=\frac{y^{2}-1}{x^{2}-1},y(2)=2$\n25. $x^{2}\frac{dy}{dx}=y-xy,y(-1)=-1$\n26. $\frac{dy}{dt}+2y = 1,y(0)=\frac{5}{2}$\n27. $sqrt{1 - y^{2}}dx-sqrt{1 - x^{2}}dy = 0,y(0)=\frac{sqrt{3}}{2}$\n28. $(1 + x^{4})dy+x(1 + 4y^{2})dx = 0,y(1)=0$\n29. $\frac{dy}{dx}=-yln y,y(0)=e$

in problems 1 - 22 solve the given differential equation by separation of variables.\n1. $\frac{dy}{dx}=sin3x$\n2. $\frac{dy}{dx}=(x + 1)^2$\n3. $dx+e^{3x}dy = 0$\n4. $dy-(y - 1)^2dx = 0$\n5. $x\frac{dy}{dx}=4y$\n6. $\frac{dy}{dx}+2xy^2 = 0$\n7. $\frac{dy}{dx}=e^{3x + 2y}$\n8. $e^{x}y\frac{dy}{dx}=e^{-y}+e^{-2x - y}$\n9. $yln x\frac{dx}{dy}=(\frac{y + 1}{x})^2$\n10. $\frac{dy}{dx}=(\frac{2y + 3}{4x + 5})^2$\n11. $cos ydx+sec^{2}xdy = 0$\n12. $sin3xdx+2ycos^{3}3xdy = 0$\n13. $(e^{x}+1)^2e^{-y}dx+(e^{x}+1)^3e^{-x}dy = 0$\n14. $x(1 + y^{2})^{1/2}dx=y(1 + x^{2})^{1/2}dy$\n15. $\frac{ds}{dr}=ks$\n16. $\frac{dq}{dt}=k(q - 70)$\n17. $\frac{dp}{dt}=p - p^{2}$\n18. $\frac{dn}{dt}+n=nte^{t + 2}$\n19. $\frac{dy}{dx}=\frac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$\n20. $\frac{dy}{dx}=\frac{xy + 2y - x - 2}{xy - 3y + x - 3}$\n21. $\frac{dy}{dx}=xsqrt{1 - y^{2}}$\n22. $(e^{x}+e^{-x})\frac{dy}{dx}=y^{2}$\nin problems 23 - 30 find an explicit solution of the given initial - value problem.\n23. $\frac{dx}{dt}=4(x^{2}+1),x(pi/4)=1$\n24. $\frac{dy}{dx}=\frac{y^{2}-1}{x^{2}-1},y(2)=2$\n25. $x^{2}\frac{dy}{dx}=y-xy,y(-1)=-1$\n26. $\frac{dy}{dt}+2y = 1,y(0)=\frac{5}{2}$\n27. $sqrt{1 - y^{2}}dx-sqrt{1 - x^{2}}dy = 0,y(0)=\frac{sqrt{3}}{2}$\n28. $(1 + x^{4})dy+x(1 + 4y^{2})dx = 0,y(1)=0$\n29. $\frac{dy}{dx}=-yln y,y(0)=e$

Answer

Answer:

We will solve problem 1 as an example: $\frac{dy}{dx}=\sin(3x)$

Separate variables: $dy = \sin(3x)dx$

Integrate both sides: $\int dy=\int\sin(3x)dx$

Let $u = 3x$, then $du=3dx$ and $dx=\frac{1}{3}du$

$\int\sin(3x)dx=\frac{1}{3}\int\sin(u)du=-\frac{1}{3}\cos(u)+C=-\frac{1}{3}\cos(3x)+C$

So $y =-\frac{1}{3}\cos(3x)+C$

Explanation:

Step1: Separate variables

$dy=\sin(3x)dx$

Step2: Integrate left - hand side

$\int dy=y$

Step3: Integrate right - hand side

Let $u = 3x$, $dx=\frac{1}{3}du$; $\int\sin(3x)dx=\frac{1}{3}\int\sin(u)du$

Step4: Evaluate integral of $\sin(u)$

$\frac{1}{3}\int\sin(u)du=-\frac{1}{3}\cos(u)+C$

Step5: Substitute back $u = 3x$

$-\frac{1}{3}\cos(3x)+C$

Step6: Write the solution

$y =-\frac{1}{3}\cos(3x)+C$