Nonhomogeneous Differential Equations — Question 1

Question 1 – Solution

Strategy. Subtract one known forced solution so that the remaining unknown satisfies a homogeneous equation.

Step 1: Prove the representation. If $L[y]=L[y_p]=g$, linearity gives $L[y-y_p]=0$. Completeness of $u,v$ then gives $y-y_p=Au+Bv$. Conversely, $L[y_p+Au+Bv]=g+A(0)+B(0)=g$. Independence makes the two constants unique for each solution. Thus $$\boxed{y=y_p+Au+Bv.}$$

Step 2: Transfer the initial data. The correction $h=y-y_p$ must satisfy $$h(t_0)=a-y_p(t_0),\qquad h'(t_0)=b-y_p'(t_0).$$ Regular homogeneous existence and uniqueness determine $h$ on $I$. Therefore the forced initial-value problem is unique too. The coefficients in the correction must fit these adjusted data, not automatically the original data.

Step 3: Verify and complete the example. For $y_p=t^2$, the left-hand side is $2-6t+2t^2$, as required. The characteristic polynomial of the associated homogeneous equation is $(r-1)(r-2)$, so $$\boxed{y=t^2+Ae^t+Be^{2t}.}$$ The exponentials are independent homogeneous solutions, and the representation theorem proves that no further solution freedom is missing.

Step 4: Fit and check. At zero, the equations are $A+B=1$ and $A+2B=0$. They give $A=2$, $B=-1$, hence $$\boxed{y=t^2+2e^t-e^{2t}.}$$ Its initial value is $2-1=1$ and its slope is $0+2-2=0$. The polynomial contributes the entire forcing, while each exponential has zero residual, verifying the differential equation everywhere.