C.9 · Integrating factors and nonhomogeneous differential equations

Undergraduate GRE preparation. Local objectives within the reviewed ETS scope; this is original teaching, not an official test or score predictor.

Prerequisites: First-order product rule, exponentials and characteristic equations.

• Solve first-order linear equations using an integrating factor
• Construct the homogeneous and particular parts of a constant-coefficient solution
• Handle resonance and verify the result in the original equation

integrating factor 积分因子: A multiplier that turns a first-order linear equation into a product derivative.

particular solution 特解: One solution supplying the specified nonhomogeneous forcing.

A first-order linear equation has the form y′+p(x)y=q(x) on an interval where its coefficients are continuous. It need not be separable. Set μ(x)=exp(∫p(x)dx). Since μ′=pμ, multiplying gives (μy)′=μq, hence y=μ⁻¹(∫μq dx+C). Choose any convenient antiderivative for p; its integration constant only rescales μ and cancels from the solution. Apply initial data after integration, and keep any singular coefficient points outside the chosen interval.

For y′+2xy=x with y(0)=1, μ=e^(x²). Then (e^(x²)y)′=x e^(x²), whose integral is e^(x²)/2+C. Thus y=1/2+C e^(−x²), and the initial condition gives C=1/2. Direct substitution checks y′+2xy=x. The integrating factor makes the left side a product derivative; it is not an extra factor that remains multiplying the original right side in the final answer.

For ay″+by′+cy=r(x) with a nonzero, solve the characteristic equation aλ²+bλ+c=0 for the homogeneous part. Distinct real roots give two exponentials; a repeated root λ gives (C₁+C₂x)e^(λx); roots α±iβ give e^(αx)(C₁ cos βx+C₂ sin βx). The complete solution is y_h+y_p. A polynomial forcing suggests a polynomial particular trial; an exponential or sine/cosine forcing suggests the corresponding family, with enough coefficients to account for differentiation.

If a particular trial duplicates a homogeneous solution, it cannot produce the forcing: multiply by x once for a simple root, twice for a repeated root. For y″−3y′+2y=e^x, the naive Ke^x is annihilated. Trying Kxe^x gives −Ke^x, so y_p=−xe^x. Initial conditions determine the homogeneous constants only after adding y_p. Substitute the final function into the original equation to check signs and forcing; an initial-value check alone cannot verify the differential equation.

For y″−3y′+2y=4, the characteristic roots are 1 and 2. A constant particular solution y_p=2 gives 2y_p=4, so y=C₁e^x+C₂e^(2x)+2. For y′+2xy=x and y(0)=1, the separate first-order solution is y=(1+e^(−x²))/2. These illustrate distinct methods; neither equation becomes homogeneous just because its left side is linear.

Keep the particular solution. A resonant trial needs an x factor. The integrating-factor derivation applies on a valid coefficient interval and should not divide by y or silently discard zero solutions.

Solve $y'+2y=e^x$, $y(0)=1$, by an integrating factor. Check the answer directly.

Worked solution

The coefficients are continuous everywhere. The integrating factor is $\mu=e^{2x}$. Multiplication gives $(e^{2x}y)'=e^{3x}$. Integration gives $y=e^x/3+Ce^{-2x}$. The initial condition yields C=2/3, so $y=e^x/3+2e^{-2x}/3$. Differentiation gives $y'=e^x/3-4e^{-2x}/3$; adding 2y gives $e^x$, and y(0)=1.

Solve $y''-2y'+y=e^x$ with $y(0)=0,y'(0)=0$. Explain why neither $Ke^x$ nor $Kxe^x$ is a sufficient particular trial.

Check after attempting

The homogeneous root is 1 with multiplicity two, so $y_h=(C_1+C_2x)e^x$. Both suggested trials already belong to that homogeneous space and are annihilated. Put $y=e^xv$; then $y''-2y'+y=e^xv''$. Therefore $v''=1$ and $v=x^2/2+C_2x+C_1$. The two initial conditions give $C_1=C_2=0$, hence $y=x^2e^x/2$. This substitution also checks the forcing without relying only on initial values.