# 3 if we let u x t v x t ψ x then we obtain as in

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3. If we let u ( x, t ) = v ( x, t ) + ψ ( x ), then we obtain as in Example 1 in the text + r = 0 or ψ ( x ) = r 2 k x 2 + c 1 x + c 2 . The boundary conditions become u (0 , t ) = v (0 , t ) + ψ (0) = u 0 u (1 , t ) = v (1 , t ) + ψ (1) = u 0 . Letting ψ (0) = ψ (1) = u 0 we obtain homogeneous boundary conditions in v : v (0 , t ) = 0 and v (1 , t ) = 0 . Now ψ (0) = ψ (1) = u 0 implies c 2 = u 0 and c 1 = r/ 2 k . Thus ψ ( x ) = r 2 k x 2 + r 2 k x + u 0 = u 0 r 2 k x ( x 1) . To determine v ( x, t ) we solve k 2 v ∂x 2 = ∂v dt , 0 < x < 1 , t > 0 v (0 , t ) = 0 , v (1 , t ) = 0 , v ( x, 0) = r 2 k x ( x 1) u 0 . Separating variables, we find v ( x, t ) = n =1 A n e kn 2 π 2 t sin nπx, where A n = 2 1 0 r 2 k x ( x 1) u 0 sin nπx dx = 2 u 0 + r kn 3 π 3 [( 1) n 1] . ( 1 ) Hence, a solution of the original problem is u ( x, t ) = ψ ( x ) + v ( x, t ) = u 0 r 2 k x ( x 1) + n =1 A n e kn 2 π 2 t sin nπx, where A n is defined in ( 1 ). 4. If we let u ( x, t ) = v ( x, t ) + ψ ( x ), then we obtain as in Example 1 in the text + r = 0 . 719

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13.6 Nonhomogeneous Equations and Boundary Conditions Integrating gives ψ ( x ) = r 2 k x 2 + c 1 x + c 2 . The boundary conditions become u (0 , t ) = v (0 , t ) + ψ (0) = u 0 u (1 , t ) = v (1 , t ) + ψ (1) = u 1 . Letting ψ (0) = u 0 and ψ (1) = u 1 we obtain homogeneous boundary conditions in v : v (0 , t ) = 0 and v (1 , t ) = 0 . Now ψ (0) = u 0 and ψ (1) = u 1 imply c 2 = u 0 and c 1 = u 1 u 0 + r/ 2 k . Thus ψ ( x ) = r 2 k x 2 + u 1 u 0 + r 2 k x + u 0 . To determine v ( x, t ) we solve k 2 v ∂x 2 = ∂v dt , 0 < x < 1 , t > 0 v (0 , t ) = 0 , v (1 , t ) = 0 , v ( x, 0) = f ( x ) ψ ( x ) . Separating variables, we find v ( x, t ) = n =1 A n e kn 2 π 2 t sin nπx, where A n = 2 1 0 [ f ( x ) ψ ( x )]sin nπx dx. ( 2 ) Hence, a solution of the original problem is u ( x, t ) = ψ ( x ) + v ( x, t ) = r 2 k x 2 + u 1 u 0 + r 2 k x + u 0 + n =1 A n e kn 2 π 2 t sin nπx, where A n is defined in ( 2 ). 5. Substituting u ( x, t ) = v ( x, t ) + ψ ( x ) into the partial differential equation gives k 2 v ∂x 2 + + Ae βx = ∂v ∂t . This equation will be homogeneous provided ψ satisfies + Ae βx = 0 . The solution of this differential equation is obtained by successive integrations: ψ ( x ) = A β 2 k e βx + c 1 x + c 2 . From ψ (0) = 0 and ψ (1) = 0 we find c 1 = A β 2 k ( e β 1) and c 2 = A β 2 k . Hence ψ ( x ) = A β 2 k e βx + A β 2 k ( e β 1) x + A β 2 k = A β 2 k 1 e βx + ( e β 1) x . 720
13.6 Nonhomogeneous Equations and Boundary Conditions Now the new problem is k 2 v ∂x 2 = ∂v ∂t , 0 < x < 1 , t > 0 , v (0 , t ) = 0 , v (1 , t ) = 0 , t > 0 , v ( x, 0) = f ( x ) ψ ( x ) , 0 < x < 1 . Identifying this as the heat equation solved in Section 13 . 3 in the text with L = 1 we obtain v ( x, t ) = n =1 A n e kn 2 π 2 t sin nπx where A n = 2 1 0 [ f ( x ) ψ ( x )]sin nπx dx. Thus u ( x, t ) = A β 2 k 1 e βx + ( e β 1) x + n =1 A n e kn 2 π 2 t sin nπx. 6. Substituting u ( x, t ) = v ( x, t ) + ψ ( x ) into the partial differential equation gives k 2 v ∂x 2 + hv = ∂v ∂t . This equation will be homogeneous provided ψ satisfies = 0 .

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