Solved on Jan 19, 2024

Solve for the general solution of the differential equation $(x^2-xy+y^2)dx-xy\,dx=0$ . If homogeneous, use $y=vx$ .

STEP 1

Assumptions
1. We are given a first-order homogeneous differential equation in the form: $(x^2 - xy + y^2)dx - xydx = 0$
2. We will use the substitution method with $y = vx$ to solve for the general solution.

STEP 2

Substitute $y = vx$ into the differential equation. Note that $dy = vdx + xdv$ because $v$ is a function of $x$ .

STEP 3

Replace $y$ with $vx$ and $dy$ with $vdx + xdv$ in the differential equation.
$(x^2 - x(vx) + (vx)^2)dx - x(vx)dx = 0$

STEP 4

Simplify the equation by distributing $x$ and combining like terms.
$(x^2 - x^2v + x^2v^2)dx - x^2vdx = 0$

STEP 5

Factor out $x^2dx$ from the terms.
$x^2dx(1 - v + v^2 - v) = 0$

STEP 6

Combine like terms within the parentheses.
$x^2dx(v^2 - 2v + 1) = 0$

STEP 7

Recognize that $v^2 - 2v + 1$ is a perfect square.
$x^2dx(v - 1)^2 = 0$

STEP 8

Since $x^2dx$ cannot be zero for the differential equation to hold, we can divide both sides by $x^2dx$ to simplify the equation.
$(v - 1)^2 = 0$

STEP 9

Solve for $v$ .
$v - 1 = 0$

STEP 10

Find the value of $v$ .
$v = 1$

STEP 11

Recall that $v = \frac{y}{x}$ , so we can now substitute back to find the relationship between $y$ and $x$ .
$\frac{y}{x} = 1$

STEP 12

Solve for $y$ .
$y = x$

STEP 13

State the general solution of the differential equation.
The general solution of the given homogeneous differential equation is $y = x$ .

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Solve for the general solution of the differential equation $(x^2-xy+y^2)dx-xy\,dx=0$ . If homogeneous, use $y=vx$ .

STEP 1

Assumptions
1. We are given a first-order homogeneous differential equation in the form: $(x^2 - xy + y^2)dx - xydx = 0$
2. We will use the substitution method with $y = vx$ to solve for the general solution.

STEP 2

Substitute $y = vx$ into the differential equation. Note that $dy = vdx + xdv$ because $v$ is a function of $x$ .

STEP 3

Replace $y$ with $vx$ and $dy$ with $vdx + xdv$ in the differential equation.
$(x^2 - x(vx) + (vx)^2)dx - x(vx)dx = 0$

STEP 4

Simplify the equation by distributing $x$ and combining like terms.
$(x^2 - x^2v + x^2v^2)dx - x^2vdx = 0$

STEP 5

Factor out $x^2dx$ from the terms.
$x^2dx(1 - v + v^2 - v) = 0$

STEP 6

Combine like terms within the parentheses.
$x^2dx(v^2 - 2v + 1) = 0$

STEP 7

Recognize that $v^2 - 2v + 1$ is a perfect square.
$x^2dx(v - 1)^2 = 0$

STEP 8

Since $x^2dx$ cannot be zero for the differential equation to hold, we can divide both sides by $x^2dx$ to simplify the equation.
$(v - 1)^2 = 0$

STEP 9

Solve for $v$ .
$v - 1 = 0$

STEP 10

Find the value of $v$ .
$v = 1$

STEP 11

Recall that $v = \frac{y}{x}$ , so we can now substitute back to find the relationship between $y$ and $x$ .
$\frac{y}{x} = 1$

STEP 12

Solve for $y$ .
$y = x$

STEP 13

State the general solution of the differential equation.
The general solution of the given homogeneous differential equation is $y = x$ .

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Solve for the general solution of the differential equation (x2−xy+y2)dx−xy dx=0(x^2-xy+y^2)dx-xy\,dx=0(x2−xy+y2)dx−xydx=0. If homogeneous, use y=vxy=vxy=vx.

STEP 1

Assumptions1. We are given a first-order homogeneous differential equation in the form: (x2−xy+y2)dx−xydx=0 (x^2 - xy + y^2)dx - xydx = 0 (x2−xy+y2)dx−xydx=02. We will use the substitution method with y=vxy = vxy=vx to solve for the general solution.

STEP 2

Substitute y=vxy = vxy=vx into the differential equation. Note that dy=vdx+xdvdy = vdx + xdvdy=vdx+xdv because vvv is a function of xxx.

STEP 3

Replace yyy with vxvxvx and dydydy with vdx+xdvvdx + xdvvdx+xdv in the differential equation.(x2−x(vx)+(vx)2)dx−x(vx)dx=0 (x^2 - x(vx) + (vx)^2)dx - x(vx)dx = 0 (x2−x(vx)+(vx)2)dx−x(vx)dx=0

STEP 4

Simplify the equation by distributing xxx and combining like terms.(x2−x2v+x2v2)dx−x2vdx=0 (x^2 - x^2v + x^2v^2)dx - x^2vdx = 0 (x2−x2v+x2v2)dx−x2vdx=0

STEP 5

Factor out x2dxx^2dxx2dx from the terms.x2dx(1−v+v2−v)=0 x^2dx(1 - v + v^2 - v) = 0 x2dx(1−v+v2−v)=0

STEP 6

Combine like terms within the parentheses.x2dx(v2−2v+1)=0 x^2dx(v^2 - 2v + 1) = 0 x2dx(v2−2v+1)=0

STEP 7

Recognize that v2−2v+1v^2 - 2v + 1v2−2v+1 is a perfect square.x2dx(v−1)2=0 x^2dx(v - 1)^2 = 0 x2dx(v−1)2=0

STEP 8

Since x2dxx^2dxx2dx cannot be zero for the differential equation to hold, we can divide both sides by x2dxx^2dxx2dx to simplify the equation.(v−1)2=0 (v - 1)^2 = 0 (v−1)2=0

STEP 9

Solve for vvv.v−1=0 v - 1 = 0 v−1=0

STEP 10

Find the value of vvv.v=1 v = 1 v=1

STEP 11

Recall that v=yxv = \frac{y}{x}v=xy​, so we can now substitute back to find the relationship between yyy and xxx.yx=1 \frac{y}{x} = 1 xy​=1

STEP 12

Solve for yyy.y=x y = x y=x

STEP 13

State the general solution of the differential equation.The general solution of the given homogeneous differential equation is y=xy = xy=x.

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Solve for the general solution of the differential equation $(x^2-xy+y^2)dx-xy\,dx=0$ . If homogeneous, use $y=vx$ .

Assumptions
1. We are given a first-order homogeneous differential equation in the form: $(x^2 - xy + y^2)dx - xydx = 0$
2. We will use the substitution method with $y = vx$ to solve for the general solution.

Substitute $y = vx$ into the differential equation. Note that $dy = vdx + xdv$ because $v$ is a function of $x$ .

Replace $y$ with $vx$ and $dy$ with $vdx + xdv$ in the differential equation.
$(x^2 - x(vx) + (vx)^2)dx - x(vx)dx = 0$

Simplify the equation by distributing $x$ and combining like terms.
$(x^2 - x^2v + x^2v^2)dx - x^2vdx = 0$

Factor out $x^2dx$ from the terms.
$x^2dx(1 - v + v^2 - v) = 0$

Combine like terms within the parentheses.
$x^2dx(v^2 - 2v + 1) = 0$

Recognize that $v^2 - 2v + 1$ is a perfect square.
$x^2dx(v - 1)^2 = 0$

Since $x^2dx$ cannot be zero for the differential equation to hold, we can divide both sides by $x^2dx$ to simplify the equation.
$(v - 1)^2 = 0$

Solve for $v$ .
$v - 1 = 0$

Find the value of $v$ .
$v = 1$

Recall that $v = \frac{y}{x}$ , so we can now substitute back to find the relationship between $y$ and $x$ .
$\frac{y}{x} = 1$

Solve for $y$ .
$y = x$

State the general solution of the differential equation.
The general solution of the given homogeneous differential equation is $y = x$ .