Solve the differential equation (x−3y)dx−xdy=0

Name: Solve the differential equation (x−3y)dx−xdy=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the differential equation (x−3y)dx−xdy=0

Question

Nathanael Webber · Accepted Answer

(x−3y)−xdydx=0 −dydx−(x−3yx)=0 dydx−xx+3yx=0 dydx+3xy=1 dydx+Py=Q where P=3x, Q=1 I.F.=e∫Pdx =e∫3xdx =e3ln⁡abs(x) =eln⁡(x)3 =x3 y(I.F.)=∫Q(I.F.)+C y(x3)=∫1(x3)+C =x44+C y=x4+Cx3 =x4+C1 C1=Cx3

Nick Camelot · Accepted Answer

To solve the differential equation (x−3y)dx−xdy=0, we can use the method of exact differential equations. The equation is exact if we can find a function u(x,y) such that the total differential du satisfies ∂u∂xdx+∂u∂ydy=0.Let&#039;s check if the given equation is exact by computing the partial derivatives:∂∂y(x−3y)=−3∂∂x(−1)=0Since the mixed partial derivative ∂∂y(−1) is not equal to ∂∂x(−3y), the equation is not exact. However, we can try to find an integrating factor to make it exact.We multiply the entire equation by an integrating factor I(x):I(x)(x−3y)dx−I(x)dy=0We want the coefficient of dy to be equal to ∂u∂y. In this case, ∂u∂y=−I(x). We can solve this equation to find u(x,y).Integrating ∂u∂y=−I(x) with respect to y, we get:u(x,y)=−I(x)y+g(x)where g(x) is an arbitrary function of x. To determine I(x), we differentiate u(x,y) with respect to x and equate it to the coefficient of dx:∂u∂x=−I′(x)y+g′(x)=x−3yComparing coefficients, we have:−I′(x)=1g′(x)=0Solving these equations, we find:I(x)=−x+Cg(x)=C′where C and C′ are constants.Now, we can substitute I(x) and g(x) back into u(x,y):u(x,y)=(−x+C)y+C′Finally, we equate u(x,y) to zero since du=0:(−x+C)y+C′=0

Eliza Beth13 · Accepted Answer

Step 1: Rewrite the equation in the standard form:

(x - 3 y) d x - x d y = 0

Step 2: Divide through by

x

to separate the variables:

\frac{(x - 3 y)}{x} d x - d y = 0

Step 3: Rearrange the terms:

\frac{d x}{x} - 3 \frac{d y}{d x} = 0

Step 4: Introduce a substitution, let

u = \frac{d y}{d x}

:

\frac{d x}{x} - 3 u d u = 0

Step 5: Integrate both sides of the equation:

\int \frac{d x}{x} - \int 3 u d u = \ln | x | - \frac{3 u^{2}}{2} = C

Step 6: Substitute back

u = \frac{d y}{d x}

:

\ln | x | - \frac{3 (\frac{d y}{d x})^{2}}{2} = C

Step 7: Solve for

y

:

\frac{d y}{d x} = \sqrt{\frac{2 (\ln | x | - C)}{3}}

Step 8: Integrate both sides with respect to

x

:

\int \frac{d y}{\sqrt{\frac{2 (\ln | x | - C)}{3}}} = \int d x

Step 9: Simplify the integral and solve for

y

:

y = \int \sqrt{\frac{3}{2}} \frac{d x}{\sqrt{\ln | x | - C}}

Therefore, the solution to the differential equation

(x - 3 y) d x - x d y = 0

is given by

y = \int \sqrt{\frac{3}{2}} \frac{d x}{\sqrt{\ln | x | - C}}

, where

C

is the constant of integration.

madeleinejames20 · Accepted Answer

Answer:x−y=3yln|x|+CExplanation:The differential equation is:(x−3y)dx−xdy=0To begin, we rearrange the terms:(x−3y)dx=xdyNow, let&#039;s divide both sides by x to separate the variables:(x−3y)xdx=dyNext, we integrate both sides with respect to their respective variables:∫(x−3y)xdx=∫dyTo integrate ∫(x−3y)xdx, we can expand the fraction:∫(1−3yx)dxIntegrating term by term, we get:∫dx−∫3yxdxIntegrating the first term with respect to x gives us x, and integrating the second term gives us −3yln|x|:x−3yln|x|=y+CHere, C represents the constant of integration.Finally, we can rearrange the equation to solve for y:x−y=3yln|x|+C

Solve the differential equation (x-3y)dx-xdy=0

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