Help With Diff Equations & Vectors?

Question

1) Solve the differential equation

dy/dx = -3x^2 + 3x + 2 + (2/x)y

2) Solve the linear differential equation

dy/dx = (4 - (x + 2)y) / (x + 1) , y(0) = 7

3) Find the scalar and vector projections of b onto a, where

b = - i - 4j + k and a = 3i + 7j + 5k

a) (scalar projection) compab =

b) (vector projection) projab =

Answer 1

dy/dx = -3x² + 3x + 2 + (2/x)·y

y' - (2/x)·y = -3x² + 3x + 2

— — — — — — — — — — — — — — — —

This is in the form:

y' + P(x)·y = f(x)

Use the integrating factor:

e^[ ∫ P(x) dx ]

→ e^[ ∫ -(2/x) dx ]

= e^[ -2 · ∫ 1/x dx ]

= e^[ -2 · ln(x) + C, ]

= e^[ -2 · ln(x) ]·e^(C,)

= { e^[ ln(x) ] }^(-2)·e^(C,)

= x^(-2)·e^(C,)

= e^(C,)/x²

— — — — — — — — — — — — — — — —

Multiply both sides by this integrating factor:

e^[ ∫ -(2/x) dx ] · [ y' - (2/x)·y ] = e^[ ∫ -(2/x) dx ] · [ -3x² + 3x + 2 ]

[ e^(C,)/x² ] · [ y' - (2/x)·y ] = [ e^(C,)/x² ] · [ -3x² + 3x + 2 ]

Which is the same as

d/dx [ ( e^(C,)/x² ) · y ] = [ e^(C,)/x² ] · [ -3x² + 3x + 2 ]

The left side is the same because that is what the integrating factor is utilized for.

Integrate both sides with respect to x:

∫ d/dx [ ( e^(C,)/x² ) · y ] dx = ∫ [ e^(C,)/x² ] · [ -3x² + 3x + 2 ] dx

( e^(C,)/x² ) · y = e^(C,) · ∫ ( 1/x² ) · ( -3x² + 3x + 2 ) dx

( 1/x² ) · y = ∫ ( 1/x² ) · ( -3x² + 3x + 2 ) dx

( 1/x² ) · y = ∫ ( -3 + 3/x + 2/x² ) dx

( 1/x² ) · y = -3·x + 3·ln(x) - 2/x + C

y = x² · ( -3·x + 3·ln(x) - 2/x + C )

You can check this in the original equation, and it comes up as a valid solution (I checked). So this family of solutions is valid.

Answer:

y(x) = x² · ( -3·x + 3·ln(x) - 2/x + C )

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For the second problem, the shorter way:

A problem of the form:

y' + P(x)·y = f(x)

Can be rewritten as:

y = e^(-∫ Pdx) · ∫ [ e^(∫Pdx) · f(x) ] dx

Thus, for the second problem:

dy/dx = (4 - (x + 2)y) / (x + 1)

y' = 4/(x + 1) - (x + 2)/(x + 1) · y

y' + (x + 2)/(x + 1) · y = 4/(x + 1)

Thus,

P(x) = (x + 2)/(x + 1)

f(x) = 4/(x + 1)

∫ Pdx → ∫(x + 2)/(x + 1)dx

Let u = x + 1

Then du = dx

→ ∫(u + 1)/(u)du

= ∫( u/u + 1/u )du

= ∫1du + ∫1/u du

= u + ln(u) + C,

= x + 1 + ln(x + 1) + C,

Combine the constants:

= x + ln(x + 1) + C,,

Put this as the power of e:

→ e^[ x + ln(x + 1) + C,, ]

= e^(x)·e^[ln(x + 1)]·e^(C,,)

= (x + 1)·e^(x)·e^(C,,)

The constant will cancel out though, so you can drop it giving:

e^(∫ Pdx) → (x + 1)·e^(x)

Now change:

y = e^(-∫ Pdx) · ∫ [ e^(∫Pdx) · f(x) ] dx

Into:

y = 1/[(x + 1)·e^(x)] · ∫ [ (x + 1)·e^(x) · 4/(x + 1) ] dx

Simplify and evaluate the integral:

y = 1/[(x + 1)·e^(x)] · ∫ [ 1·e^(x) · 4/1 ] dx

y = 4/[(x + 1)·e^(x)] · ∫ e^(x) dx

y = 4/[(x + 1)·e^(x)] · [ e^(x) + C,,, ]

This can simplify to:

y = 4/(x + 1) + 4·C,,,/[(x + 1)·e^(x) ]

Four times a constant is just another constant, giving the final result:

y = 4/(x + 1) + C/[ (x + 1)·e^(x) ]

You can check this family of solutions by substituting it into the original problem (again, I did).

Now, to use the initial condition:

y(0) = 7 = 4/(0 + 1) + C/[ (0 + 1)·e^(0) ]

7 = 4/(1) + C/[ (0 + 1)·(1) ]

7 = 4 + C

3 = C

The specific solution is:

y(x) = 4/(x + 1) + 3/[ (x + 1)·e^(x) ]