What is the solution to solving a linear differential equation?

[ariffinaldo]

Homework Statement

1. To use Euler midpoint method to find an approx value to y(0.1) using a step size 0.1
2. To use integrating factor method to find exact value of y(0.1)
3. Determine the global error ( min 5 decimal places)

Homework Equations

f(x,y) = x+y, y(0) = 1.35

The Attempt at a Solution

I have done euler midpoint part which is
y1 = y0 + hf(x0, y0)= 1.35 +0.1(0+0.1) = 1.45
Is my working and answer correct. Pls any help..

 

[HallsofIvy]

Homework Statement

1. To use Euler midpoint method to find an approx value to y(0.1) using a step size 0.1
2. To use integrating factor method to find exact value of y(0.1)
3. Determine the global error ( min 5 decimal places)

Homework Equations

f(x,y) = x+y, y(0) = 1.35

The Attempt at a Solution

I have done euler midpoint part which is
y1 = y0 + hf(x0, y0)= 1.35 +0.1(0+0.1)

This is incorrect. It should be y(0)+ h(x+ y)= 1.35+ .1(0+ 1.35), not "(0+ .1)"

= 1.45
Is my working and answer correct. Pls any help..

 

[ariffinaldo]

This is incorrect. It should be y(0)+ h(x+ y)= 1.35+ .1(0+ 1.35), not "(0+ .1)"

Ohh yes.. how could i miss that? How abt the intergrating factor part.. I am a loss for that.

 

[HallsofIvy]

The differential equation y'= x+y or y'+ y= x is linear. You should have learned a formula for the integrating factor for any linear d.e.

If not, you can work it out as follow: an integrating factor is a function u(x) such that multiplying the equation by it makes the left side a single derivative:
u(x)(y'+ y)= u(x)y'+ u(x)y= (uy)'. By the product rule, (uy)'= uy'+ u'y so that will be equal to uy'+ uy if and only if u'= u. What function satifies that?

Once you have found u, and multiplied by it, your equation will be (uy)'= ux. Integrating that gives
u(x)y(x)= \int u(x)x dx+ C
(The "u(x)" on the right is inside the integration. You cannot just divide both sides by u(x) and eliminate that.)