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Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation. x²y3 + x(5 + y²)y' = 0₁ µ(x, y) = Here, M(x, y) = x²y³ and N(x, y) = x dx + . Therefore, My = dy = 0. Now we see that for this equation M = x and N = 4y = h'(y) = N = Therefore, h(y) = and Nx= . We see that the equation is not exact. Now, multiplying the equation by µ(x, y) = = Therefore, M = = N₁. Integrating M with respect to x, we see that y = and we conclude that the solution of the equation is given implicitly by and y = 0. the equation becomes + h(y). Further,

Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation. x²y3 + x(5 + y²)y' = 0₁ µ(x, y) = Here, M(x, y) = x²y³ and N(x, y) = x dx + . Therefore, My = dy = 0. Now we see that for this equation M = x and N = 4y = h'(y) = N = Therefore, h(y) = and Nx= . We see that the equation is not exact. Now, multiplying the equation by µ(x, y) = = Therefore, M = = N₁. Integrating M with respect to x, we see that y = and we conclude that the solution of the equation is given implicitly by and y = 0. the equation becomes + h(y). Further,

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 33CR
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Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation. x²y3 + x(5 + y²)y' = 0₁ µ(x, y) = Here, M(x, y) = x²y³ and N(x, y) = x dx + . Therefore, My = dy = 0. Now we see that for this equation M = x and N = 4y = h'(y) = N = Therefore, h(y) = and Nx= . We see that the equation is not exact. Now, multiplying the equation by µ(x, y) = = Therefore, M = = N₁. Integrating M with respect to x, we see that y = and we conclude that the solution of the equation is given implicitly by and y = 0. the equation becomes + h(y). Further,
Transcribed Image Text:Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation. x²y3 + x(5 + y²)y' = 0₁ µ(x, y) = Here, M(x, y) = x²y³ and N(x, y) = x dx + . Therefore, My = dy = 0. Now we see that for this equation M = x and N = 4y = h'(y) = N = Therefore, h(y) = and Nx= . We see that the equation is not exact. Now, multiplying the equation by µ(x, y) = = Therefore, M = = N₁. Integrating M with respect to x, we see that y = and we conclude that the solution of the equation is given implicitly by and y = 0. the equation becomes + h(y). Further,
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