2.6 Differential equations

Write the equation as dy/dx+P(x)y=Q(x). The integrating factor is μ=e^{∫P(x)dx}; multiplying through makes the left side d(μy)/dx, so μy=∫μQ dx+C.

Put every y term on the left and divide by the coefficient of dy/dx before choosing μ. Apply an initial condition only after the general solution is obtained.

For dy/dx+2y=e^x, μ=e^{2x}; then d(e^{2x}y)/dx=e^{3x}, so y=(1/3)e^x+Ce^{−2x}.

The integrating factor multiplies the whole equation, not just y, and P must be read after standardising the derivative coefficient.

The general solution of a linear ODE is complementary function (the homogeneous equation) plus a particular integral (one solution for the forcing term). Initial conditions determine the constants.

For constant coefficients, solve the auxiliary equation for the complementary function. Choose a particular form matching the forcing, multiplying by x when it duplicates a complementary term.

For y″−3y′+2y=e^x, the auxiliary roots are 1 and 2; because e^x duplicates a complementary term, try Axe^x for the particular solution.

The complementary function alone cannot satisfy a non-zero forcing term, and a guessed particular form must be adjusted when resonance occurs.

For ay″+by′+cy=0, try y=e^{mx}; the auxiliary equation am²+bm+c=0 determines the complementary function. Distinct real roots give two exponentials, a repeated root gives (C₁+C₂x)e^{mx}, and complex roots give e^{αx}(C₁cosβx+C₂sinβx).

The form follows from the root type, not from memorised labels. Substitute the proposed solution back into the differential equation to check the signs and coefficients.

For y″−4y′+13y=0, m²−4m+13=0 gives m=2±3i, so y=e^{2x}(C₁cos3x+C₂sin3x).

Complex roots do not make y complex when real initial data are used; the conjugate pair combines into real sine and cosine terms.

For ay″+by′+cy=F(x), write y=yc+yp. The complementary function yc solves the homogeneous equation; a particular integral yp accounts for the forcing term.

Choose a trial form matching F(x): exponentials, polynomials, sines/cosines or their combinations. If the trial duplicates part of yc, multiply by enough powers of x to make it independent.

For y″+y=cos x, cos x and sin x already belong to yc, so a trial such as Ax sin x is required rather than A cos x+B sin x.

The complementary function alone solves only the zero-forcing equation, and a trial form that duplicates yc cannot determine a particular solution.

Choose a new variable u=g(y) or u=g(x,y) so that the derivatives in the equation combine into du/dx or a simpler separable form. The substitution is useful only when it reduces the order or nonlinearity.

Differentiate the substitution explicitly, replace every occurrence consistently and recover y only after solving for u. Check any lost solutions introduced by division or square roots.

For an equation containing y′/y, set u=ln|y| so u′=y′/y; the transformed equation may become linear or directly integrable.

A substitution is not just a change of symbol: forgetting the chain-rule factor produces a different differential equation.

Solving an ODE gives a family containing arbitrary constants. Initial conditions such as y(0)=a and y′(0)=b provide simultaneous equations that determine those constants.

Keep the general solution and its derivative visible, substitute the conditions at the same point, and solve the resulting linear system. For a second-order equation, two independent conditions are normally needed.

If y=C₁e^x+C₂e^{2x}, y(0)=3 gives C₁+C₂=3; y′(0)=4 gives C₁+2C₂=4, so C₁=2 and C₂=1.

One condition cannot determine two arbitrary constants, and applying a condition before differentiating can lose necessary information.