ANSWER on Question #79307 – Math – Differential Equations
QUESTION
Exact equations
( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x + ( x 3 ) ( e x 2 y ) d y = 0 (e^{x^2}y)(1 + 2(x^2)y)dx + (x^3)(e^{x^2}y)dy = 0 ( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x + ( x 3 ) ( e x 2 y ) d y = 0
Integrating factor:
1. Differential equation
( 2 x y ) d x + ( 2 ( x 2 ) + 3 ) d y = 0 (2xy)dx + (2(x^2) + 3)dy = 0 ( 2 x y ) d x + ( 2 ( x 2 ) + 3 ) d y = 0
Having an integration factor is just a function of y y y
2. Differential equation
( ( x 2 ) + 3 x + 2 ) d x + ( ( x 2 ) + x + 1 ) d y = 0 ((x^2) + 3x + 2)dx + ((x^2) + x + 1)dy = 0 (( x 2 ) + 3 x + 2 ) d x + (( x 2 ) + x + 1 ) d y = 0
Having an integration factor is just a function of ( x + y ) (x + y) ( x + y )
SOLUTION
The first part: solve the above differential equation.
1. STEP: We transform this differential equation
( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x + ( x 3 ) ( e x 2 y ) d y = 0 → ( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x = − ( x 3 ) ( e x 2 y ) d y ∣ ÷ ( − 1 d x ⋅ ( x 3 ) ( e x 2 y ) ) → − ( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x d x ⋅ ( x 3 ) ( e x 2 y ) = ( x 3 ) ( e x 2 y ) d y d x ⋅ ( x 3 ) ( e x 2 y ) → − ( 1 + 2 ( x 2 ) y ) x 3 = d y d x → d y d x = − 1 x 3 − 2 y x 2 x 3 → d y d x = − 1 x 3 − 2 y x → d y d x + 2 y x = − 1 x 3 \begin{aligned}
(e^{x^2}y)(1 + 2(x^2)y)dx + (x^3)(e^{x^2}y)dy &= 0 \rightarrow \\
(e^{x^2}y)(1 + 2(x^2)y)dx &= -(x^3)(e^{x^2}y)dy \Big| \div \left( \frac{-1}{dx \cdot (x^3)(e^{x^2}y)} \right) \rightarrow \\
&\quad - \frac{(e^{x^2}y)(1 + 2(x^2)y)dx}{dx \cdot (x^3)(e^{x^2}y)} = \frac{(x^3)(e^{x^2}y)dy}{dx \cdot (x^3)(e^{x^2}y)} \rightarrow -\frac{(1 + 2(x^2)y)}{x^3} = \frac{dy}{dx} \rightarrow \\
&\quad \frac{dy}{dx} = -\frac{1}{x^3} - \frac{2yx^2}{x^3} \rightarrow \frac{dy}{dx} = -\frac{1}{x^3} - \frac{2y}{x} \rightarrow \boxed{\frac{dy}{dx} + \frac{2y}{x} = -\frac{1}{x^3}}
\end{aligned} ( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x + ( x 3 ) ( e x 2 y ) d y ( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x = 0 → = − ( x 3 ) ( e x 2 y ) d y ∣ ∣ ÷ ( d x ⋅ ( x 3 ) ( e x 2 y ) − 1 ) → − d x ⋅ ( x 3 ) ( e x 2 y ) ( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x = d x ⋅ ( x 3 ) ( e x 2 y ) ( x 3 ) ( e x 2 y ) d y → − x 3 ( 1 + 2 ( x 2 ) y ) = d x d y → d x d y = − x 3 1 − x 3 2 y x 2 → d x d y = − x 3 1 − x 2 y → d x d y + x 2 y = − x 3 1
Conclusion,
( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x + ( x 3 ) ( e x 2 y ) d y = 0 → d y d x + 2 y x = − 1 x 3 (e^{x^2}y)(1 + 2(x^2)y)dx + (x^3)(e^{x^2}y)dy = 0 \rightarrow \frac{dy}{dx} + \frac{2y}{x} = -\frac{1}{x^3} ( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x + ( x 3 ) ( e x 2 y ) d y = 0 → d x d y + x 2 y = − x 3 1
Nonhomogeneous differential equation of the first order.
2 STEP: Let us solve the transformed equation.
Since,
d y d x + 2 y x = − 1 x 3 \frac{dy}{dx} + \frac{2y}{x} = -\frac{1}{x^3} d x d y + x 2 y = − x 3 1
Nonhomogeneous differential equation of the first order, then the solution consists of two parts:
y ( x ) = y 1 ( x ) + y 2 ( x ) , where { y 1 ( x ) − solution of the homogeneous equation y 2 ( x ) − a particular solution of the nonhomogeneous equation y(x) = y_1(x) + y_2(x), \quad \text{where} \quad \left\{ \begin{array}{c} y_1(x) - \text{solution of the homogeneous equation} \\ y_2(x) - a \text{ particular solution of the nonhomogeneous equation} \end{array} \right. y ( x ) = y 1 ( x ) + y 2 ( x ) , where { y 1 ( x ) − solution of the homogeneous equation y 2 ( x ) − a particular solution of the nonhomogeneous equation
2A STEP: We solve the homogeneous equation.
For the solution, we use the method of separation of variables.
(More information: https://en.wikipedia.org/wiki/Separation_of_variables)
d y d x + 2 y x = 0 → d y d x = − 2 y x → d y y = − 2 d x x → ∫ d y y = ∫ ( − 2 d x x ) → ln ∣ y ∣ = − 2 ⋅ ln ∣ x ∣ + ln ∣ C ∣ → \frac{dy}{dx} + \frac{2y}{x} = 0 \rightarrow \frac{dy}{dx} = -\frac{2y}{x} \rightarrow \frac{dy}{y} = -2\frac{dx}{x} \rightarrow \int \frac{dy}{y} = \int \left(-2\frac{dx}{x}\right) \rightarrow \ln |y| = -2 \cdot \ln |x| + \ln |C| \rightarrow d x d y + x 2 y = 0 → d x d y = − x 2 y → y d y = − 2 x d x → ∫ y d y = ∫ ( − 2 x d x ) → ln ∣ y ∣ = − 2 ⋅ ln ∣ x ∣ + ln ∣ C ∣ → ln ∣ y ∣ = ln ∣ x − 2 ∣ + ln ∣ C ∣ → ln ∣ y ∣ = ln ∣ C ⋅ x − 2 ∣ → y 1 ( x ) = C x 2 \ln |y| = \ln |x^{-2}| + \ln |C| \rightarrow \ln |y| = \ln |C \cdot x^{-2}| \rightarrow y_1(x) = \frac{C}{x^2} ln ∣ y ∣ = ln ∣ x − 2 ∣ + ln ∣ C ∣ → ln ∣ y ∣ = ln ∣ C ⋅ x − 2 ∣ → y 1 ( x ) = x 2 C
Conclusion,
d y d x + 2 y x = 0 → y 1 ( x ) = C x 2 \boxed{\frac{dy}{dx} + \frac{2y}{x} = 0 \rightarrow y_1(x) = \frac{C}{x^2}} d x d y + x 2 y = 0 → y 1 ( x ) = x 2 C
2B STEP: We solve the nonhomogeneous equation.
For the solution, we use the method of variation of the parameter.
(More information: https://en.wikipedia.org/wiki/Variation_of_parameters)
\left\{ \begin{array}{c} \frac{dy}{dx} + \frac{2y}{x} = -\frac{1}{x^3} \\ y(x) = \frac{C(x)}{x^2} \rightarrow \frac{dy}{dx} = \frac{1}{x^2} \cdot \frac{dC}{dx} - \frac{2C(x)}{x^2} \rightarrow \frac{1}{x^2} \cdot \frac{dC}{dx} - \frac{2C(x)}{x^2} + \frac{2}{x} \cdot \frac{C(x)}{x^2} = -\frac{1}{x^3} \rightarrow \\ \frac{1}{x^2} \cdot \frac{dC}{dx} = -\frac{1}{x^3} \right| \times (dx \cdot x^2) \rightarrow dC = -\frac{dx}{x} \rightarrow \int dC = \int \left(-\frac{dx}{x}\right) \rightarrow \boxed{C(x) = -\ln |x| + C_1}
Then,
{ y ( x ) = C ( x ) x 2 C ( x ) = C 1 − ln ∣ x ∣ → y ( x ) = C 1 − ln ∣ x ∣ x 2 = C 1 x 2 − ln ∣ x ∣ x , where { y 1 ( x ) = C 1 x 2 y 2 ( x ) = − ln ∣ x ∣ x \left\{ \begin{array}{c} y (x) = \frac {C (x)}{x ^ {2}} \\ C (x) = C _ {1} - \ln | x | \end{array} \right. \to y (x) = \frac {C _ {1} - \ln | x |}{x ^ {2}} = \frac {C _ {1}}{x ^ {2}} - \frac {\ln | x |}{x}, \text{ where } \left\{ \begin{array}{c} y _ {1} (x) = \frac {C _ {1}}{x ^ {2}} \\ y _ {2} (x) = - \frac {\ln | x |}{x} \end{array} \right. { y ( x ) = x 2 C ( x ) C ( x ) = C 1 − ln ∣ x ∣ → y ( x ) = x 2 C 1 − ln ∣ x ∣ = x 2 C 1 − x ln ∣ x ∣ , where { y 1 ( x ) = x 2 C 1 y 2 ( x ) = − x l n ∣ x ∣
Conclusion,
( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x + ( x 3 ) ( e x 2 y ) d y = 0 → y ( x ) = C 1 x 2 − ln ∣ x ∣ x \boxed{(e ^ {x ^ {2}} y) (1 + 2 (x ^ {2}) y) d x + (x ^ {3}) (e ^ {x ^ {2}} y) d y = 0 \rightarrow y (x) = \frac {C _ {1}}{x ^ {2}} - \frac {\ln | x |}{x}} ( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x + ( x 3 ) ( e x 2 y ) d y = 0 → y ( x ) = x 2 C 1 − x ln ∣ x ∣
The second part: we solve the problems associated with the integral factor.
We recall the definition of an equation in exact differentials:
M ( x , y ) d x + N ( x , y ) d y = 0 M(x, y)dx + N(x, y)dy = 0 M ( x , y ) d x + N ( x , y ) d y = 0
is exact if
∂ M ∂ y = ∂ N ∂ x \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} ∂ y ∂ M = ∂ x ∂ N
Recall another definition:
If the equation
M ( x , y ) d x + N ( x , y ) d y = 0 M(x, y)dx + N(x, y)dy = 0 M ( x , y ) d x + N ( x , y ) d y = 0
is not exact, but the equation
μ ( x , y ) M ( x , y ) d x + μ ( x , y ) N ( x , y ) d y = 0 \mu(x, y)M(x, y)dx + \mu(x, y)N(x, y)dy = 0 μ ( x , y ) M ( x , y ) d x + μ ( x , y ) N ( x , y ) d y = 0
is exact, then μ ( x , y ) \mu(x, y) μ ( x , y )
We recall one theorem connected with the integral factor.
Theorem:
If
∂ M / ∂ y − ∂ N / ∂ x N is continuous and depends only on x , then \frac{\partial M / \partial y - \partial N / \partial x}{N} \text{ is continuous and depends only on } x, \text{ then} N ∂ M / ∂ y − ∂ N / ∂ x is continuous and depends only on x , then μ ( x ) = exp [ ∫ ( ∂ M / ∂ y − ∂ N / ∂ x N ) d x ] is an integrating factor for the D E . \mu(x) = \exp \left[ \int \left( \frac{\partial M / \partial y - \partial N / \partial x}{N} \right) dx \right] \text{ is an integrating factor for the } DE. μ ( x ) = exp [ ∫ ( N ∂ M / ∂ y − ∂ N / ∂ x ) d x ] is an integrating factor for the D E .
If
∂ N / ∂ x − ∂ M / ∂ y M is continuous and depends only on y , then \frac{\partial N / \partial x - \partial M / \partial y}{M} \text{ is continuous and depends only on } y, \text{ then} M ∂ N / ∂ x − ∂ M / ∂ y is continuous and depends only on y , then μ ( y ) = exp [ ∫ ( ∂ N / ∂ x − ∂ M / ∂ y M ) d y ] is an integrating factor for the D E . \mu(y) = \exp \left[ \int \left( \frac{\partial N / \partial x - \partial M / \partial y}{M} \right) dy \right] \text{ is an integrating factor for the } DE. μ ( y ) = exp [ ∫ ( M ∂ N / ∂ x − ∂ M / ∂ y ) d y ] is an integrating factor for the D E .
Now we can start solving the problem.
1. Differential equation
( 2 x y ) d x + ( 2 ( x 2 ) + 3 ) d y = 0 (2xy) \, dx + (2(x^2) + 3) \, dy = 0 ( 2 x y ) d x + ( 2 ( x 2 ) + 3 ) d y = 0
Having an integration factor is just a function of y y y
In our case,
{ M ( x , y ) = 2 x y N ( x , y ) = 2 x 2 + 3 → { ∂ M ∂ y = 2 x ∂ N ∂ x = 2 ⋅ 2 x = 4 x → ∂ M ∂ y = 2 x ≠ 4 x = ∂ N ∂ x \left\{
\begin{array}{l}
M(x, y) = 2xy \\
N(x, y) = 2x^2 + 3
\end{array}
\right.
\rightarrow
\left\{
\begin{array}{l}
\frac{\partial M}{\partial y} = 2x \\
\frac{\partial N}{\partial x} = 2 \cdot 2x = 4x
\end{array}
\right.
\rightarrow
\frac{\partial M}{\partial y} = 2x \neq 4x = \frac{\partial N}{\partial x} { M ( x , y ) = 2 x y N ( x , y ) = 2 x 2 + 3 → { ∂ y ∂ M = 2 x ∂ x ∂ N = 2 ⋅ 2 x = 4 x → ∂ y ∂ M = 2 x = 4 x = ∂ x ∂ N
Conclusion,
( 2 x y ) d x + ( 2 ( x 2 ) + 3 ) d y = 0 is not exact (2xy) \, dx + (2(x^2) + 3) \, dy = 0 \text{ is not exact} ( 2 x y ) d x + ( 2 ( x 2 ) + 3 ) d y = 0 is not exact
We now verify the fulfillment of the conditions of the theorem:
∂ N / ∂ x − ∂ M / ∂ y M = 4 x − 2 x 2 x y = 2 x 2 x y = 1 y is continuous and depends only on y \frac{\partial N / \partial x - \partial M / \partial y}{M} = \frac{4x - 2x}{2xy} = \frac{2x}{2xy} = \frac{1}{y} \text{ is continuous and depends only on } y M ∂ N / ∂ x − ∂ M / ∂ y = 2 x y 4 x − 2 x = 2 x y 2 x = y 1 is continuous and depends only on y
Then,
∃ μ ( y ) is an integrating factor for the DE . \exists \mu(y) \text{ is an integrating factor for the DE}. ∃ μ ( y ) is an integrating factor for the DE .
We can easily find this integral factor, using the formula from the theorem
μ ( y ) = exp [ ∫ ( ∂ N / ∂ x − ∂ M / ∂ y M ) d y ] = exp [ ∫ ( 1 y ) d y ] = exp ( ln y ) = y → μ ( y ) = y \mu(y) = \exp \left[ \int \left(\frac{\partial N / \partial x - \partial M / \partial y}{M}\right) dy \right] = \exp \left[ \int \left(\frac{1}{y}\right) dy \right] = \exp(\ln y) = y \rightarrow \boxed{\mu(y) = y} μ ( y ) = exp [ ∫ ( M ∂ N / ∂ x − ∂ M / ∂ y ) d y ] = exp [ ∫ ( y 1 ) d y ] = exp ( ln y ) = y → μ ( y ) = y
Checking:
( 2 x y ) d x + ( 2 ( x 2 ) + 3 ) d y = 0 ∣ × ( y ) → ( 2 x y 2 ) d x + ( 2 y x 2 + 3 y ) d y = 0 → { M ( x , y ) = 2 x y 2 N ( x , y ) = ( 2 x 2 y + 3 y ) → { ∂ M ∂ y = 4 x y ∂ N ∂ x = 2 ⋅ 2 x y = 4 x y → ∂ M ∂ y = 4 x y = ∂ N ∂ x \begin{array}{l}
(2xy) \, dx + (2(x^2) + 3) \, dy = 0 \mid \times (y) \rightarrow (2xy^2) \, dx + (2yx^2 + 3y) \, dy = 0 \rightarrow \\
\left\{
\begin{array}{l}
M(x, y) = 2xy^2 \\
N(x, y) = (2x^2y + 3y)
\end{array}
\right.
\rightarrow
\left\{
\begin{array}{l}
\frac{\partial M}{\partial y} = 4xy \\
\frac{\partial N}{\partial x} = 2 \cdot 2xy = 4xy
\end{array}
\right.
\rightarrow
\frac{\partial M}{\partial y} = 4xy = \frac{\partial N}{\partial x} \\
\end{array} ( 2 x y ) d x + ( 2 ( x 2 ) + 3 ) d y = 0 ∣ × ( y ) → ( 2 x y 2 ) d x + ( 2 y x 2 + 3 y ) d y = 0 → { M ( x , y ) = 2 x y 2 N ( x , y ) = ( 2 x 2 y + 3 y ) → { ∂ y ∂ M = 4 x y ∂ x ∂ N = 2 ⋅ 2 x y = 4 x y → ∂ y ∂ M = 4 x y = ∂ x ∂ N
Conclusion,
( 2 x y 2 ) d x + ( 2 y x 2 + 3 y ) d y = 0 is exact (2xy^2) \, dx + (2yx^2 + 3y) \, dy = 0 \text{ is exact} ( 2 x y 2 ) d x + ( 2 y x 2 + 3 y ) d y = 0 is exact
2. Differential equation
( ( x 2 ) + 3 x + 2 ) d x + ( ( x 2 ) + x + 1 ) d y = 0 ((x^2) + 3x + 2)dx + ((x^2) + x + 1)dy = 0 (( x 2 ) + 3 x + 2 ) d x + (( x 2 ) + x + 1 ) d y = 0
Having an integration factor is just a function of ( x + y ) (x + y) ( x + y )
In our case,
{ M ( x , y ) = x 2 + 3 x + 2 N ( x , y ) = x 2 + x + 1 → { ∂ M ∂ y = 0 ∂ N ∂ x = 2 x + 1 → ∂ M ∂ y = 0 ≠ 2 x + 1 = ∂ N ∂ x \left\{ \begin{array}{l}
M(x, y) = x^2 + 3x + 2 \\
N(x, y) = x^2 + x + 1
\end{array} \right.
\rightarrow
\left\{ \begin{array}{l}
\frac{\partial M}{\partial y} = 0 \\
\frac{\partial N}{\partial x} = 2x + 1
\end{array} \right.
\rightarrow
\frac{\partial M}{\partial y} = 0 \neq 2x + 1 = \frac{\partial N}{\partial x} { M ( x , y ) = x 2 + 3 x + 2 N ( x , y ) = x 2 + x + 1 → { ∂ y ∂ M = 0 ∂ x ∂ N = 2 x + 1 → ∂ y ∂ M = 0 = 2 x + 1 = ∂ x ∂ N
Conclusion,
( x 2 ) + 3 x + 2 ⋅ d x + ( ( x 2 ) + x + 1 ) d y = 0 is not exact \boxed{(x^2) + 3x + 2 \cdot dx + ((x^2) + x + 1)dy = 0 \text{ is not exact}} ( x 2 ) + 3 x + 2 ⋅ d x + (( x 2 ) + x + 1 ) d y = 0 is not exact
We now verify the fulfillment of the conditions of the theorem:
∂ M / ∂ y − ∂ N / ∂ x N = 0 − ( 2 x + 1 ) x 2 + x + 1 = − 2 x + 1 x 2 + x + 1 is continuous and depends only on x \frac{\partial M / \partial y - \partial N / \partial x}{N} = \frac{0 - (2x + 1)}{x^2 + x + 1} = -\frac{2x + 1}{x^2 + x + 1} \text{ is continuous and depends only on } x N ∂ M / ∂ y − ∂ N / ∂ x = x 2 + x + 1 0 − ( 2 x + 1 ) = − x 2 + x + 1 2 x + 1 is continuous and depends only on x
Then,
∃ μ ( x ) is an integrating factor for the DE . \exists \mu(x) \text{ is an integrating factor for the DE}. ∃ μ ( x ) is an integrating factor for the DE .
We can easily find this integral factor, using the formula from the theorem
μ ( x ) = exp [ ∫ ( ∂ M / ∂ y − ∂ N / ∂ x N ) d x ] = exp [ ∫ ( − 2 x + 1 x 2 + x + 1 ) d x ] = [ x 2 + x + 1 = t ( 2 x + 1 ) d x = d t ] = = exp [ ∫ ( − 1 t ) d t ] = exp [ − ln ∣ t ∣ ] = exp [ ln ∣ 1 t ∣ ] = 1 t = 1 x 2 + x + 1 → μ ( x ) = 1 x 2 + x + 1 \begin{aligned}
\mu(x) &= \exp \left[ \int \left(\frac{\partial M / \partial y - \partial N / \partial x}{N}\right) dx \right] = \exp \left[ \int \left(-\frac{2x + 1}{x^2 + x + 1}\right) dx \right] = \begin{bmatrix} x^2 + x + 1 &= t \\ (2x + 1)dx &= dt \end{bmatrix} = \\
&= \exp \left[ \int \left(-\frac{1}{t}\right) dt \right] = \exp \left[ -\ln |t| \right] = \exp \left[ \ln \left| \frac{1}{t} \right| \right] = \frac{1}{t} = \frac{1}{x^2 + x + 1} \rightarrow \boxed{ \mu(x) = \frac{1}{x^2 + x + 1} }
\end{aligned} μ ( x ) = exp [ ∫ ( N ∂ M / ∂ y − ∂ N / ∂ x ) d x ] = exp [ ∫ ( − x 2 + x + 1 2 x + 1 ) d x ] = [ x 2 + x + 1 ( 2 x + 1 ) d x = t = d t ] = = exp [ ∫ ( − t 1 ) d t ] = exp [ − ln ∣ t ∣ ] = exp [ ln ∣ ∣ t 1 ∣ ∣ ] = t 1 = x 2 + x + 1 1 → μ ( x ) = x 2 + x + 1 1
Checking:
( ( x 2 ) + 3 x + 2 ) d x + ( ( x 2 ) + x + 1 ) d y = 0 ∣ × ( 1 x 2 + x + 1 ) → \left(\left(x ^ {2}\right) + 3 x + 2\right) d x + \left(\left(x ^ {2}\right) + x + 1\right) d y = 0 \mid \times \left(\frac {1}{x ^ {2} + x + 1}\right)\rightarrow ( ( x 2 ) + 3 x + 2 ) d x + ( ( x 2 ) + x + 1 ) d y = 0 ∣ × ( x 2 + x + 1 1 ) → ( x 2 + 3 x + 2 ) x 2 + x + 1 d x + ( x 2 + x + 1 ) x 2 + x + 1 d y = 0 → ( x 2 + 3 x + 2 x 2 + x + 1 ) ⋅ d x + 1 ⋅ d y = 0 \frac {(x ^ {2} + 3 x + 2)}{x ^ {2} + x + 1} d x + \frac {(x ^ {2} + x + 1)}{x ^ {2} + x + 1} d y = 0 \rightarrow \left(\frac {x ^ {2} + 3 x + 2}{x ^ {2} + x + 1}\right) \cdot d x + 1 \cdot d y = 0 x 2 + x + 1 ( x 2 + 3 x + 2 ) d x + x 2 + x + 1 ( x 2 + x + 1 ) d y = 0 → ( x 2 + x + 1 x 2 + 3 x + 2 ) ⋅ d x + 1 ⋅ d y = 0 { M ( x , y ) = x 2 + 3 x + 2 x 2 + x + 1 N ( x , y ) = 1 → { ∂ M ∂ y = 0 ∂ N ∂ x = 0 → ∂ M ∂ y = 0 = ∂ N ∂ x \left\{ \begin{array}{c} M (x, y) = \frac {x ^ {2} + 3 x + 2}{x ^ {2} + x + 1} \\ N (x, y) = 1 \end{array} \right. \to \left\{ \begin{array}{l} \frac {\partial M}{\partial y} = 0 \\ \frac {\partial N}{\partial x} = 0 \end{array} \right. \to \frac {\partial M}{\partial y} = 0 = \frac {\partial N}{\partial x} { M ( x , y ) = x 2 + x + 1 x 2 + 3 x + 2 N ( x , y ) = 1 → { ∂ y ∂ M = 0 ∂ x ∂ N = 0 → ∂ y ∂ M = 0 = ∂ x ∂ N
Conclusion,
( x 2 + 3 x + 2 x 2 + x + 1 ) ⋅ d x + 1 ⋅ d y = 0 is exact \boxed {\left(\frac {x ^ {2} + 3 x + 2}{x ^ {2} + x + 1}\right) \cdot d x + 1 \cdot d y = 0 \text{ is exact}} ( x 2 + x + 1 x 2 + 3 x + 2 ) ⋅ d x + 1 ⋅ d y = 0 is exact
ANSWER:
( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x + ( x 3 ) ( e x 2 y ) d y = 0 → y ( x ) = C 1 x 2 − ln ∣ x ∣ x \left(e ^ {x ^ {2}} y\right) (1 + 2 (x ^ {2}) y) d x + (x ^ {3}) \left(e ^ {x ^ {2}} y\right) d y = 0 \rightarrow y (x) = \frac {C _ {1}}{x ^ {2}} - \frac {\ln | x |}{x} ( e x 2 y ) ( 1 + 2 ( x 2 ) y ) d x + ( x 3 ) ( e x 2 y ) d y = 0 → y ( x ) = x 2 C 1 − x ln ∣ x ∣
1.
μ ( y ) = y is an integrating factor for the DE \mu (y) = y \text{ is an integrating factor for the DE} μ ( y ) = y is an integrating factor for the DE μ ( y ) = y is a function only of the variable y \mu (y) = y \text{ is a function only of the variable } y μ ( y ) = y is a function only of the variable y
2.
μ ( x ) = x 2 + x + 1 is an integrating factor for the DE \mu (x) = x ^ {2} + x + 1 \text{ is an integrating factor for the DE} μ ( x ) = x 2 + x + 1 is an integrating factor for the DE μ ( x ) = x 2 + x + 1 is not a function of a variable ( x + y ) \mu (x) = x ^ {2} + x + 1 \text{ is not a function of a variable } (x + y) μ ( x ) = x 2 + x + 1 is not a function of a variable ( x + y )
Answer provided by https://www.AssignmentExpert.com
#340153 on Dec 2023