General and particular solutions for 3 differential equations

Question

Hello, I've got some problems solving general and particular solutions of differential equations. 

I hope to understand how they work, if you could help me with a detailed solution for these 3 exercises:

 

A) x² dy/dx = y² , particular solution for y=2,when x=1

B) x dy/dx = (1+x)/cos y,  y=π/2, when x=1

C) dy/dx= (1-2x)cos²y,  y=π/4, when x=0

Answer 1

A) Rearrange equation  (divide both sides by x² and by y²)to make

dy/y²=dx/x²

 

Integrating both sides gives the equation

 

y^-1=x^-1 + C, where C is a constant.  This is the general solution.

 

when x=1, y=2, we can find a value for C.

 

2^-1=1^-1 + C

 

1/2=1 + C

 

-1/2=C

 

y^-1=x^-1 -1/2.  This is the particular solution because this value of C holds true for these values of x and y.

 

B)

 

Rearrange the equation to get

 

cos(y)*dy=[(1+x)/x]*dx

 

cos(y)*dy=(1/x +1)*dx

 

Integrate both sides to get

 

sin(y)=ln(x) + x + C.  This is the general solution.

 

Find C when y=n/2, x=1

 

sin(n/2)=ln(1) + 1 + C

 

sin(n/2)=1 + C

 

sin(n/2) - 1=C

 

sin(y)=ln(x) + x + sin(n/2) - 1.  This is the particular solution.

 

C)

Rearrange equation to make:

 

dy/cos²(y)=(1-2x)*dx

 

Integrate both sides to get

 

tan(y)=x -x² + C.  This is the general solution.

 

when x=0, y=n/4, find value of C.

 

tan(n/4)=0 - 0 + C.

 

tan(n/4)=C

 

tan(y)=x - x² + tan(n/4).  Particular Solution

Answer 2

(A) dy/y^2 = dx/x^2

     integral dy/y^2 = integral dx/x^2

     -1/y = - 1/x + C

     for y = 2 when x = 1, we have -1/2 = - 1 + C

     then C = 1/2

     The solution - 1/y = - 1/x + 1/2

     y = (1-2x)/x

 

(B) x dy/dx = (1+x)cos y

      integral cos y dy = integral (1+x)/x dx

      sin y = ln x + x + C

      for y = pi/2, when x = 1, we have 1 = 0 + 1 + C

      then C = 0

      the solution sin y = ln x + x

      y = arc sin (ln x + x)

 

(C)  dy/dx = (1 - 2x) cos^2 y

      integral dy/cos^2 y = integral (1 - 2x) dx

      tan y = x - x^2 + C

      for y = pi/4, when x = 0, we have 1 = 0 - 0 + C

      then C = 1

      The solution tan y = x - x^2 + 1

      y = arc tan (x - x^2 + 1)