Variable Separation, 6

Category: Differential Equations, Integral Calculus

"Published in Newark, California, USA"

Find the general solution for 

Solution:

Consider the given equation above

Transpose xy to the right side of the equation, we have

Arrange the above equation by separation of variables, we have

Integrate on both sides of the equation, we have

Take the inverse natural logarithm on both sides of the equation

where K = e^C. Therefore, the general solution is

 

Solving 3rd Order Differential Equations

Category: Differential Equations, Integral Calculus

"Published in Newark, California, USA"

Find the general solution for

Solution:

Consider the given equation above

The given equation is a 3rd Order Differential Equation because the third derivative of y with respect to x is involved. We can rewrite given equation as follows

Multiply both sides of the equation by dx, we have

Integrate on both sides of the equation, we have

Rewrite the above equation as follows

Multiply both sides of the equation by dx, we have

Integrate on both sides of the equation, we have

Multiply both sides of the equation by dx, we have

Integrate on both sides of the equation, we have

where B = ½ C₁. Therefore, the general solution is

      

Solving 2nd Order Differential Equations

Category: Differential Equations, Integral Calculus

"Published in Newark, California, USA"

Find the general solution for

Solution:

Consider the given equation above

The given equation is a 2nd Order Differential Equation because the second derivative of y with respect to x is involved. We can rewrite given equation as follows

Multiply both sides of the equation by dx, we have

Integrate on both sides of the equation, we have

Multiply both sides of the equation by dx, we have

Integrate on both sides of the equation, we have

but

Hence, the above equation becomes

where B = C₁ - 1.Therefore, the general solution is