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Problem:
Solve the following differential equations by first finding an integrating factor. $$ (y^2(x+1) + y ) \, dx + ( 2xy + 1 ) \, dy = 0 $$
Answer:
\begin{align*} M_y &= 2(x+1)y + 1 = 2xy + 2y + 1 \\ N_x &= 2y \\ \frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &= \frac{ 2xy + 2y + 1 - 2y } { 2xy + 1 } \\ \frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &= \frac{ 2xy + 1 } { 2xy + 1 } = 1 \\ \end{align*} This means that: $$ e ^ { \int \frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] \, dx } $$ is the integrating factor we seek. Call this integrating factor $I$. \begin{align*} I &= e ^ { \int 1 \, dx } = e^x \\ (y^2(x+1) + y ) e^x \, dx + ( 2xy + 1 ) e^x \, dy &= 0 \end{align*} Now we have: \begin{align*} M &= (y^2(x+1) + y ) e^x \\ M_y &= ( 2(x+1)y + 1 )e^x = ( 2xy + 2y + 1)e^x \\ N &= ( 2xy + 1 ) e^x \\ N_x &= ( 2xy + 1 ) e^x \end{align*} As I understand it, I was suppose to get $M_y = N_x$. That is, the de should have been exact. What did I do wrong?

Now, I have an updated answer. However, It is still wrong. I feel I am much closer to the right answer. Here is my updated answer:

\begin{align*} M_y &= 2(x+1)y + 1 = 2xy + 2y + 1 \\ N_x &= 2y \\ \frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &= \frac{ 2xy + 2y + 1 - 2y } { 2xy + 1 } \\ \frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &= \frac{ 2xy + 1 } { 2xy + 1 } = 1 \\ \end{align*} This means that: $$ e ^ { \int \frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] \, dx } $$ is the integrating factor we seek. Call this integrating factor $I$. \begin{align*} I &= e ^ { \int 1 \, dx } = e^x \\ (y^2(x+1) + y ) e^x \, dx + ( 2xy + 1 ) e^x \, dy &= 0 \end{align*} Now we have: \begin{align*} M &= (y^2(x+1) + y ) e^x \\ M_y &= ( 2(x+1)y + 1 )e^x = ( 2xy + 2y + 1)e^x \\ N &= ( 2xy + 1 ) e^x \\ N_x &= ( 2xy + 1 ) e^x + (2y)e^x = (2xy + 2y + 2)e^x1 \end{align*} Hence the differential equation is exact. We have: \begin{align*} F_x &= (y^2(x+1) + y ) e^x \\ F &= \int (y^2(x+1) + y ) e^x \, dx = \int (x y^2 + y^2 + 1 ) e^x \, dx \end{align*} Recall that: $$ \int x e^x \, dx = x e^x - e^x + C $$ \begin{align*} F &= y^2 \int xe^x \, dx + (y^2+1) \int e^x \, dx \\ F &= y^2 ( xe^x - e^x) + (y^2 + 1)e^x + \phi(y) \\ F &= y^2 xe^x - y^2 e^x + y^2 e^x + e^x + \phi(y) \\ F &= y^2 xe^x + e^x + \phi(y) \\ F_y &= 2xy e^x + \phi'(y) \\ 2xy e^x + \phi'(y) &= ( 2xy + 1 ) e^x \\ \phi'(y) &= e^x \\ \phi(y) &= ye^x + c \\ F &= y^2 xe^x + e^x + ye^x + c \end{align*} However, the book gets: $$ x y^2 e^x + y e^x = c $$ Where did I go wrong?

Problem:

Solve the following differential equations first finding an integrating factor. $$ ( 5xy + 4y^2 + 1 ) \, dx + ( x^2 + 2xy ) \, dy = 0 $$

Answer:

Now, I try $x^3$ as an integrating factor. This gives me: $$ ( 5x^4 y + 4 x^3 y^2 + x^3 ) \, dx + ( x^5 + 2x^4 y ) \, dy = 0 $$ Now, we see if it is exact. \begin{align*} M_y &= 5x^4 + 8 x^3 y \\ N_x &= 5x^4 + 8 x^3 y \end{align*} The equation is exact. Let $F$ be the solution we seek: \begin{align*} F_x &= 5x^4 y + 4 x^3 y^2 + x^3 \\ F &= x^5 y + x^4 y^2 + \frac{x^4}{4} + \phi(y) \\ F_y &= 5x^4 + 2x^4 y + \phi'(y) = x^5 + 2x^4 y \\ \phi'(y) &= 0 \\ \phi(y &= C \end{align*} Hence the solution we seek is: $$ 4x^5 y + 4x^4 y^2 + x^4 + C = 0 $$

Where did I go wrong?

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2 Answers 2

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$$(y^2(x+1) + y ) \, dx + ( 2xy + 1 ) \, dy = 0$$ Rearrange some terms: $$y^2xdx+y^2dx + y dx + xdy^2 + dy = 0$$ $$y^2xdx+(y^2dx +xdy^2)+ y dx + dy = 0$$ $$y^2xdx+dxy^2+ y dx + dy = 0$$ Multiply by $e^x$: $$y^2xde^x+e^xdxy^2+ y de^x + e^x dy = 0$$ $$dxy^2e^x+ de^xy= 0$$ Integrate: $$xy^2e^x+ e^xy= C$$


Note that you have to use the product rule for $N_x$: $$N_x = (( 2xy + 1 ) e^x)'$$ $$N_x= ( 2xy + 1 ) e^x+e^x(2y)$$ $$N_x= ( 2xy + 1 +2y) e^x$$

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Everything looks good up to the last formula, you missed there one factor in the product rule for $N_x$: $$ N_x=(2xy+1)_xe^x+(2xy+1)(e^x)_x=(2y)e^x+(2xy+1)e^x, $$ restoring the needed equality $M_y=N_x$.

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