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Math EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

(a) To Find: A condition on M and N that is necessary and sufficient for M d x + N d y = 0 to have an integrating factor that depends only on the product x 2 y . The required condition is H ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y ( M x 2 − 2 N x ) depends only on x 2 y _ . Theorem used: Exact equation: “Suppose the first partial derivatives of M ( x , y ) and N ( x , y ) are continuous in a rectangle R . Then M ( x , y ) d x + N ( x , y ) d y = 0 is an exact equation in R if and only if the compatibility condition ∂ M ∂ y ( x , y ) = ∂ N ∂ x ( x , y ) holds for all ( x , y ) in R .” Calculation: The differential equation is, M d x + N d y = 0 ... ( 1 ) Let μ ( x 2 y ) be the integrating factor that depends only on the product x 2 y . Multiply the equation (1) by μ ( x 2 y ) . μ ( x 2 y ) M d x + μ ( x 2 y ) N d y = 0 ... ( 2 ) Here, ∂ F ∂ x = μ ( x 2 y ) M and ∂ F ∂ y = μ ( x 2 y ) N . Differentiate ∂ F ∂ x with respect to y as shown below. ∂ ∂ y ( ∂ F ∂ x ) = ∂ ∂ y ( μ ( x 2 y ) M ) = μ ( x 2 y ) ∂ M ∂ y + x 2 ∂ μ ( x 2 y ) ∂ ( x 2 y ) ⋅ ∂ ( x 2 y ) ∂ y = μ ( x 2 y ) ∂ M ∂ y + x 2 μ ′ ( x 2 y ) M = μ ( x 2 y ) ∂ M ∂ y + M x 2 μ ′ ( x 2 y ) Differentiate ∂ F ∂ y with respect to x as shown below. ∂ ∂ x ( ∂ F ∂ y ) = ∂ ∂ x ( μ ( x 2 y ) N ) = μ ( x 2 y ) ∂ N ∂ x + N ∂ μ ( x 2 y ) ∂ ( x 2 y ) ⋅ ∂ ( x 2 y ) ∂ x = μ ( x 2 y ) ∂ N ∂ x + N μ ′ ( x 2 y ) ⋅ 2 x = μ ( x 2 y ) ∂ N ∂ x + 2 N x μ ′ ( x 2 y ) The given equation (2) to be exact, then ∂ ∂ y ( ∂ F ∂ x ) = ∂ ∂ x ( ∂ F ∂ y ) . μ ( x 2 y ) ∂ M ∂ y + M x 2 μ ′ ( x 2 y ) = μ ( x 2 y ) ∂ N ∂ x + 2 N x μ ′ ( x 2 y ) M x 2 μ ′ ( x 2 y ) − 2 N x μ ′ ( x 2 y ) = μ ( x 2 y ) ∂ N ∂ x − μ ( x 2 y ) ∂ M ∂ y ( M x 2 − 2 N x ) μ ′ ( x 2 y ) = μ ( x 2 y ) ( ∂ N ∂ x − ∂ M ∂ y ) μ ′ ( x 2 y ) μ ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y ( M x 2 − 2 N x ) Take, μ ′ ( x 2 y ) μ ( x 2 y ) = H ( x 2 y ) . The above result becomes H ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y ( M x 2 − 2 N x ) Hence, the required condition is H ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y ( M x 2 − 2 N x ) depends only on x 2 y _ . (b) To solve: The equation ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 by using an answer to part (a). The solution of the differential equation ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 is e x 2 y ( x 2 + 2 x y ) = C _ . Procedure used: “a) If M ( x , y ) d x + N ( x , y ) d y = 0 is exact, then ∂ F ∂ x = M . Integrate the last equation with respect to x to get F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) b) To determine g ( y ) , take the partial derivative with respect to y of both sides of equation F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) and substitute N for ∂ F ∂ y . We can now solve for g ′ ( y ) . c) Integrate g ′ ( y ) to obtain g ( y ) up to a numerical constant. Substituting g ( y ) into equation F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) gives F ( x , y ) . d) The solution to M d x + N d y = 0 is given implicitly by F ( x , y ) = C .” Calculation: From part (a), the necessary and sufficient condition for M d x + N d y = 0 to have an integrating factor that depends only on the product x 2 y is H ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y x ( M x − 2 N y ) where H ( x 2 y ) = μ ′ ( x 2 y ) μ ( x 2 y ) . Find the general formula for the integrating factor as follows. μ ′ ( x 2 y ) μ ( x 2 y ) = H ( x 2 y ) Integrate on both sides of the equation. ∫ μ ′ ( x 2 y ) μ ( x 2 y ) = ∫ H ( x 2 y ) d ( x 2 y ) ln μ ( x 2 y ) = ∫ H ( x 2 y ) d ( x 2 y ) μ ( x 2 y ) = exp ( ∫ H ( x 2 y ) d ( x 2 y ) ) μ ( z ) = e ∫ H ( z ) d z where z = x 2 y Thus, the general formula for the integrating factor is μ ( z ) = e ∫ H ( z ) d z where z = x 2 y . The differential equation is, ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 ... ( 1 ) Compare ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 with the equation M ( x , y ) d x + N ( x , y ) d y = 0 and the following is obtained. M ( x , y ) = 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 N ( x , y ) = 2 x + x 4 + 2 x 3 y Differentiate M ( x , y ) with respect to y as shown below. ∂ M ∂ y = ∂ ∂ y ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) = 2 + 2 x 3 + 8 x 2 y Differentiate N ( x , y ) with respect to x as shown below. ∂ N ∂ x = ∂ ∂ x ( 2 x + x 4 + 2 x 3 y ) = 2 + 4 x 3 + 6 x 2 y Calculate H ( x 2 y ) as follows. H ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y ( M x 2 − 2 N x y ) = 2 + 4 x 3 + 6 x 2 y − ( 2 + 2 x 3 + 8 x 2 y ) x 2 ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) − 2 x y ( 2 x + x 4 + 2 x 3 y ) = 2 + 4 x 3 + 6 x 2 y − 2 − 2 x 3 − 8 x 2 y 2 x 3 + 2 x 2 y + 2 x 5 y + 4 x 4 y 2 − 4 x 2 y − 2 x 5 y − 4 x 4 y 2 = 2 x 3 − 2 x 2 y 2 x 3 − 2 x 2 y = 1 Therefore, the function H ( x 2 y ) = 1 . Find the integrating factor μ ( x 2 y ) as follows. μ ( z ) = e ∫ H ( z ) d z ( where z = x 2 y ) μ ( z ) = e ∫ ( 1 ) d z μ ( z ) = e z μ ( x 2 y ) = e x 2 y ( substitute back , z = x 2 y ) Multiply the equation (1) by f ( x 2 y ) , to make the equation is exact e x 2 y [ ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y ] = 0 e x 2 y ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + e x 2 y ( 2 x + x 4 + 2 x 3 y ) d y = 0 Integrate N ( x , y ) with respect to y and obtain F ( x , y ) as shown below. F ( x , y ) = ∫ ( 2 x + x 4 + 2 x 3 y ) e x 2 y d y + g ( x ) = ∫ 2 x e x 2 y d y + ∫ x 4 e x 2 y d y + 2 x 3 ∫ y ︸ u e x 2 y d y ︸ d v + g ( x ) = 2 x e x 2 y x 2 + x 4 e x 2 y x 2 + 2 x 3 ( ( y ) ( e x 2 y x 2 ) − ∫ e x 2 y x 2 d y ) + g ( x ) = 2 e x 2 y x + x 2 e x 2 y + 2 x y e x 2 y − 2 x 3 ( e x 2 y x 4 ) + g ( x ) Simplify further, F ( x , y ) = 2 e x 2 y x + x 2 e x 2 y + 2 x y e x 2 y − 2 e x 2 y x + g ( x ) = x 2 e x 2 y + 2 x y e x 2 y + g ( x ) Differentiate F ( x , y ) = x 2 e x 2 y + 2 x y e x 2 y + g ( x ) partially with respect to x and substitute 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 for M as shown below. ∂ F ∂ x = M ∂ F ∂ x ( x 2 e x 2 y + 2 x y e x 2 y + g ( x ) ) = ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) e x 2 y 2 x e x 2 y + x 2 ⋅ 2 x y e x 2 y + 2 y e x 2 y + 2 x y ⋅ 2 x y e x 2 y + g ′ ( x ) = ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) e x 2 y ( 2 x + 2 x 3 y + 2 y + 4 x 2 y 2 ) e x 2 y + g ′ ( x ) = ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) e x 2 y g ′ ( x ) = 0 Note that as the choice of constant of integration is not important obtain g ( x ) as follows. g ( x ) = ∫ ( 0 ) d x = 0 Substitute g ( x ) = 0 in F ( x , y ) = x 2 e x 2 y + 2 x y e x 2 y + g ( x ) and the following is obtained. F ( x , y ) = x 2 e x 2 y + 2 x y e x 2 y Obtain the solution of the differential equation ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 implicitly as shown below. x 2 e x 2 y + 2 x y e x 2 y = C e x 2 y ( x 2 + 2 x y ) = C Hence, the solution of the differential equation ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 is e x 2 y ( x 2 + 2 x y ) = C _ .

(a) To Find: A condition on M and N that is necessary and sufficient for M d x + N d y = 0 to have an integrating factor that depends only on the product x 2 y . The required condition is H ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y ( M x 2 − 2 N x ) depends only on x 2 y _ . Theorem used: Exact equation: “Suppose the first partial derivatives of M ( x , y ) and N ( x , y ) are continuous in a rectangle R . Then M ( x , y ) d x + N ( x , y ) d y = 0 is an exact equation in R if and only if the compatibility condition ∂ M ∂ y ( x , y ) = ∂ N ∂ x ( x , y ) holds for all ( x , y ) in R .” Calculation: The differential equation is, M d x + N d y = 0 ... ( 1 ) Let μ ( x 2 y ) be the integrating factor that depends only on the product x 2 y . Multiply the equation (1) by μ ( x 2 y ) . μ ( x 2 y ) M d x + μ ( x 2 y ) N d y = 0 ... ( 2 ) Here, ∂ F ∂ x = μ ( x 2 y ) M and ∂ F ∂ y = μ ( x 2 y ) N . Differentiate ∂ F ∂ x with respect to y as shown below. ∂ ∂ y ( ∂ F ∂ x ) = ∂ ∂ y ( μ ( x 2 y ) M ) = μ ( x 2 y ) ∂ M ∂ y + x 2 ∂ μ ( x 2 y ) ∂ ( x 2 y ) ⋅ ∂ ( x 2 y ) ∂ y = μ ( x 2 y ) ∂ M ∂ y + x 2 μ ′ ( x 2 y ) M = μ ( x 2 y ) ∂ M ∂ y + M x 2 μ ′ ( x 2 y ) Differentiate ∂ F ∂ y with respect to x as shown below. ∂ ∂ x ( ∂ F ∂ y ) = ∂ ∂ x ( μ ( x 2 y ) N ) = μ ( x 2 y ) ∂ N ∂ x + N ∂ μ ( x 2 y ) ∂ ( x 2 y ) ⋅ ∂ ( x 2 y ) ∂ x = μ ( x 2 y ) ∂ N ∂ x + N μ ′ ( x 2 y ) ⋅ 2 x = μ ( x 2 y ) ∂ N ∂ x + 2 N x μ ′ ( x 2 y ) The given equation (2) to be exact, then ∂ ∂ y ( ∂ F ∂ x ) = ∂ ∂ x ( ∂ F ∂ y ) . μ ( x 2 y ) ∂ M ∂ y + M x 2 μ ′ ( x 2 y ) = μ ( x 2 y ) ∂ N ∂ x + 2 N x μ ′ ( x 2 y ) M x 2 μ ′ ( x 2 y ) − 2 N x μ ′ ( x 2 y ) = μ ( x 2 y ) ∂ N ∂ x − μ ( x 2 y ) ∂ M ∂ y ( M x 2 − 2 N x ) μ ′ ( x 2 y ) = μ ( x 2 y ) ( ∂ N ∂ x − ∂ M ∂ y ) μ ′ ( x 2 y ) μ ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y ( M x 2 − 2 N x ) Take, μ ′ ( x 2 y ) μ ( x 2 y ) = H ( x 2 y ) . The above result becomes H ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y ( M x 2 − 2 N x ) Hence, the required condition is H ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y ( M x 2 − 2 N x ) depends only on x 2 y _ . (b) To solve: The equation ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 by using an answer to part (a). The solution of the differential equation ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 is e x 2 y ( x 2 + 2 x y ) = C _ . Procedure used: “a) If M ( x , y ) d x + N ( x , y ) d y = 0 is exact, then ∂ F ∂ x = M . Integrate the last equation with respect to x to get F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) b) To determine g ( y ) , take the partial derivative with respect to y of both sides of equation F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) and substitute N for ∂ F ∂ y . We can now solve for g ′ ( y ) . c) Integrate g ′ ( y ) to obtain g ( y ) up to a numerical constant. Substituting g ( y ) into equation F ( x , y ) = ∫ M ( x , y ) d x + g ( y ) gives F ( x , y ) . d) The solution to M d x + N d y = 0 is given implicitly by F ( x , y ) = C .” Calculation: From part (a), the necessary and sufficient condition for M d x + N d y = 0 to have an integrating factor that depends only on the product x 2 y is H ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y x ( M x − 2 N y ) where H ( x 2 y ) = μ ′ ( x 2 y ) μ ( x 2 y ) . Find the general formula for the integrating factor as follows. μ ′ ( x 2 y ) μ ( x 2 y ) = H ( x 2 y ) Integrate on both sides of the equation. ∫ μ ′ ( x 2 y ) μ ( x 2 y ) = ∫ H ( x 2 y ) d ( x 2 y ) ln μ ( x 2 y ) = ∫ H ( x 2 y ) d ( x 2 y ) μ ( x 2 y ) = exp ( ∫ H ( x 2 y ) d ( x 2 y ) ) μ ( z ) = e ∫ H ( z ) d z where z = x 2 y Thus, the general formula for the integrating factor is μ ( z ) = e ∫ H ( z ) d z where z = x 2 y . The differential equation is, ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 ... ( 1 ) Compare ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 with the equation M ( x , y ) d x + N ( x , y ) d y = 0 and the following is obtained. M ( x , y ) = 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 N ( x , y ) = 2 x + x 4 + 2 x 3 y Differentiate M ( x , y ) with respect to y as shown below. ∂ M ∂ y = ∂ ∂ y ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) = 2 + 2 x 3 + 8 x 2 y Differentiate N ( x , y ) with respect to x as shown below. ∂ N ∂ x = ∂ ∂ x ( 2 x + x 4 + 2 x 3 y ) = 2 + 4 x 3 + 6 x 2 y Calculate H ( x 2 y ) as follows. H ( x 2 y ) = ∂ N ∂ x − ∂ M ∂ y ( M x 2 − 2 N x y ) = 2 + 4 x 3 + 6 x 2 y − ( 2 + 2 x 3 + 8 x 2 y ) x 2 ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) − 2 x y ( 2 x + x 4 + 2 x 3 y ) = 2 + 4 x 3 + 6 x 2 y − 2 − 2 x 3 − 8 x 2 y 2 x 3 + 2 x 2 y + 2 x 5 y + 4 x 4 y 2 − 4 x 2 y − 2 x 5 y − 4 x 4 y 2 = 2 x 3 − 2 x 2 y 2 x 3 − 2 x 2 y = 1 Therefore, the function H ( x 2 y ) = 1 . Find the integrating factor μ ( x 2 y ) as follows. μ ( z ) = e ∫ H ( z ) d z ( where z = x 2 y ) μ ( z ) = e ∫ ( 1 ) d z μ ( z ) = e z μ ( x 2 y ) = e x 2 y ( substitute back , z = x 2 y ) Multiply the equation (1) by f ( x 2 y ) , to make the equation is exact e x 2 y [ ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y ] = 0 e x 2 y ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + e x 2 y ( 2 x + x 4 + 2 x 3 y ) d y = 0 Integrate N ( x , y ) with respect to y and obtain F ( x , y ) as shown below. F ( x , y ) = ∫ ( 2 x + x 4 + 2 x 3 y ) e x 2 y d y + g ( x ) = ∫ 2 x e x 2 y d y + ∫ x 4 e x 2 y d y + 2 x 3 ∫ y ︸ u e x 2 y d y ︸ d v + g ( x ) = 2 x e x 2 y x 2 + x 4 e x 2 y x 2 + 2 x 3 ( ( y ) ( e x 2 y x 2 ) − ∫ e x 2 y x 2 d y ) + g ( x ) = 2 e x 2 y x + x 2 e x 2 y + 2 x y e x 2 y − 2 x 3 ( e x 2 y x 4 ) + g ( x ) Simplify further, F ( x , y ) = 2 e x 2 y x + x 2 e x 2 y + 2 x y e x 2 y − 2 e x 2 y x + g ( x ) = x 2 e x 2 y + 2 x y e x 2 y + g ( x ) Differentiate F ( x , y ) = x 2 e x 2 y + 2 x y e x 2 y + g ( x ) partially with respect to x and substitute 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 for M as shown below. ∂ F ∂ x = M ∂ F ∂ x ( x 2 e x 2 y + 2 x y e x 2 y + g ( x ) ) = ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) e x 2 y 2 x e x 2 y + x 2 ⋅ 2 x y e x 2 y + 2 y e x 2 y + 2 x y ⋅ 2 x y e x 2 y + g ′ ( x ) = ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) e x 2 y ( 2 x + 2 x 3 y + 2 y + 4 x 2 y 2 ) e x 2 y + g ′ ( x ) = ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) e x 2 y g ′ ( x ) = 0 Note that as the choice of constant of integration is not important obtain g ( x ) as follows. g ( x ) = ∫ ( 0 ) d x = 0 Substitute g ( x ) = 0 in F ( x , y ) = x 2 e x 2 y + 2 x y e x 2 y + g ( x ) and the following is obtained. F ( x , y ) = x 2 e x 2 y + 2 x y e x 2 y Obtain the solution of the differential equation ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 implicitly as shown below. x 2 e x 2 y + 2 x y e x 2 y = C e x 2 y ( x 2 + 2 x y ) = C Hence, the solution of the differential equation ( 2 x + 2 y + 2 x 3 y + 4 x 2 y 2 ) d x + ( 2 x + x 4 + 2 x 3 y ) d y = 0 is e x 2 y ( x 2 + 2 x y ) = C _ .

Solution Summary: The author explains how to find a condition that is necessary and sufficient for Mdx+Ndy=0 to have an integrating factor that depends only on the product.

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

9th Edition

ISBN: 9780135963777

Author: Nagle

Publisher: PEARSON CO

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EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 716, solve the equation. 7. xdydx=1y3 Ch. 2.2 - In Problems 716, solve the equation. 8. dxdt=3xt2 Ch. 2.2 - In Problems 716, solve the equation. 9....Ch. 2.2 - In Problems 716, solve the equation. 10....

Ch. 2.2 - In Problems 716, solve the equation. 11....Ch. 2.2 - In Problems 716, solve the equation. 12....Ch. 2.2 - In Problems 716, solve the equation. 13....Ch. 2.2 - In Problems 716, solve the equation. 14. dxdtx3=x Ch. 2.2 - In Problems 716, solve the equation. 15....Ch. 2.2 - In Problems 716, solve the equation. 16. y1 dy +...Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 23E Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 27E Ch. 2.2 - Sketch the solution to the initial value problem...Ch. 2.2 - Uniqueness Questions. In Chapter 1 we indicated...Ch. 2.2 - As stated in this section, the separation of...Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Mixing. Suppose a brine containing 0.3 kilogram...Ch. 2.2 - Newtons Law of Cooling. According to Newtons law...Ch. 2.2 - Prob. 35E Ch. 2.2 - Prob. 36E Ch. 2.2 - Compound Interest. If P(t) is the amount of...Ch. 2.2 - Free Fall. In Section 2.1, we discussed a model...Ch. 2.2 - Grand Prix Race. Driver A had been leading...Ch. 2.2 - Prob. 40E Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - Radioactive Decay. In Example 2 assume that the...Ch. 2.3 - Prob. 24E Ch. 2.3 - (a) Using definite integration, show that the...Ch. 2.3 - Prob. 26E Ch. 2.3 - Constant Multiples of Solutions. (a) Show that y =...Ch. 2.3 - Prob. 29E Ch. 2.3 - Bernoulli Equations. The equation (18) dydx+2y=xy2...Ch. 2.3 - Prob. 31E Ch. 2.3 - Prob. 32E Ch. 2.3 - Prob. 33E Ch. 2.3 - Prob. 34E Ch. 2.3 - Prob. 35E Ch. 2.3 - Prob. 36E Ch. 2.3 - Prob. 37E Ch. 2.3 - Prob. 38E Ch. 2.3 - Prob. 39E Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - Prob. 9E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 15E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 17E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 25E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 27E Ch. 2.4 - For each of the following equations, find the most...Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Orthogonal Trajectories. A geometric problem...Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.5 - Prob. 1E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 3E Ch. 2.5 - Prob. 4E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 6E Ch. 2.5 - Prob. 7E Ch. 2.5 - Prob. 8E Ch. 2.5 - Prob. 9E Ch. 2.5 - Prob. 10E Ch. 2.5 - Prob. 11E Ch. 2.5 - Prob. 12E Ch. 2.5 - Prob. 13E Ch. 2.5 - Prob. 14E Ch. 2.5 - Prob. 15E Ch. 2.5 - Prob. 16E Ch. 2.5 - Prob. 17E Ch. 2.5 - Prob. 18E Ch. 2.5 - Prob. 19E Ch. 2.5 - Verify that when the linear differential equation...Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 2E Ch. 2.6 - Prob. 3E Ch. 2.6 - Prob. 4E Ch. 2.6 - Prob. 5E Ch. 2.6 - Prob. 6E Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 8E Ch. 2.6 - Use the method discussed under Homogeneous...Ch. 2.6 - Prob. 10E Ch. 2.6 - Prob. 11E Ch. 2.6 - Prob. 12E Ch. 2.6 - Prob. 13E Ch. 2.6 - Prob. 14E Ch. 2.6 - Prob. 15E Ch. 2.6 - Prob. 16E Ch. 2.6 - Prob. 17E Ch. 2.6 - Prob. 18E Ch. 2.6 - Prob. 19E Ch. 2.6 - Prob. 20E Ch. 2.6 - Prob. 21E Ch. 2.6 - Prob. 22E Ch. 2.6 - Use the method discussed under Bernoulli Equations...Ch. 2.6 - Prob. 24E Ch. 2.6 - Prob. 25E Ch. 2.6 - Prob. 26E Ch. 2.6 - Prob. 27E Ch. 2.6 - Prob. 28E Ch. 2.6 - Use the method discussed under Equations with...Ch. 2.6 - Prob. 30E Ch. 2.6 - Prob. 31E Ch. 2.6 - Prob. 32E Ch. 2.6 - Prob. 33E Ch. 2.6 - Prob. 34E Ch. 2.6 - Prob. 35E Ch. 2.6 - In Problems 3340, solve the equation given in: 36....Ch. 2.6 - Prob. 37E Ch. 2.6 - Prob. 38E Ch. 2.6 - Prob. 39E Ch. 2.6 - Prob. 40E Ch. 2.6 - Prob. 41E Ch. 2.6 - Prob. 42E Ch. 2.6 - Prob. 43E Ch. 2.6 - Show that equation (13) reduces to an equation of...Ch. 2.6 - Prob. 45E Ch. 2.6 - Prob. 46E Ch. 2.6 - Prob. 47E Ch. 2.6 - Prob. 48E Ch. 2 - In Problems 130, solve the equation. 1....Ch. 2 - Prob. 2RP Ch. 2 - Prob. 3RP Ch. 2 - Prob. 4RP Ch. 2 - Prob. 5RP Ch. 2 - In Problems 130, solve the equation. 6. 2xy3 dx ...Ch. 2 - In Problems 130, solve the equation. 7. t3y2 dt +...Ch. 2 - Prob. 8RP Ch. 2 - In Problems 130, solve the equation. 9. (x2 + y2)...Ch. 2 - Prob. 10RP Ch. 2 - Prob. 11RP Ch. 2 - Prob. 12RP Ch. 2 - Prob. 13RP Ch. 2 - Prob. 14RP Ch. 2 - Prob. 15RP Ch. 2 - Prob. 16RP Ch. 2 - Prob. 17RP Ch. 2 - Prob. 18RP Ch. 2 - Prob. 19RP Ch. 2 - Prob. 20RP Ch. 2 - Prob. 21RP Ch. 2 - Prob. 22RP Ch. 2 - Prob. 23RP Ch. 2 - In Problems 130, solve the equation. 24. (y/x +...Ch. 2 - Prob. 25RP Ch. 2 - Prob. 26RP Ch. 2 - Prob. 27RP Ch. 2 - Prob. 28RP Ch. 2 - Prob. 29RP Ch. 2 - Prob. 30RP Ch. 2 - Prob. 31RP Ch. 2 - Prob. 32RP Ch. 2 - Prob. 33RP Ch. 2 - Prob. 34RP Ch. 2 - Prob. 35RP Ch. 2 - Prob. 36RP Ch. 2 - Prob. 37RP Ch. 2 - Prob. 38RP Ch. 2 - Prob. 39RP Ch. 2 - Prob. 40RP Ch. 2 - Prob. 41RP

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