Solved on Mar 28, 2024

Solve the ordinary differential equation y3x+(4y+3x)dydx=0y-3x + (4y+3x)\frac{dy}{dx}=0 using the homogeneous method.

STEP 1

1. The given equation is a first-order ordinary differential equation (ODE).
2. The equation can be solved using the homogeneous method, which implies that it can be rewritten in the form M(x,y)dx+N(x,y)dy=0M(x, y)dx + N(x, y)dy = 0, where MM and NN are homogeneous functions of the same degree.
3. A function f(x,y)f(x, y) is homogeneous of degree nn if f(tx,ty)=tnf(x,y)f(tx, ty) = t^n f(x, y) for all tt.

STEP 2

1. Rewrite the equation in the form M(x,y)dx+N(x,y)dy=0M(x, y)dx + N(x, y)dy = 0.
2. Verify that M(x,y)M(x, y) and N(x,y)N(x, y) are homogeneous functions of the same degree.
3. Use the substitution y=vxy = vx (or x=vyx = vy) to transform the ODE into a separable equation.
4. Solve the resulting separable ODE.
5. Substitute back to find the solution in terms of xx and yy.

STEP 3

Rewrite the given equation in the form M(x,y)dx+N(x,y)dy=0M(x, y)dx + N(x, y)dy = 0.
y3x+(4y+3x)dydx=0 y - 3x + (4y + 3x)\frac{dy}{dx} = 0 (4y+3x)dydx=3xy \Rightarrow (4y + 3x)\frac{dy}{dx} = 3x - y 3xdx+4ydx=3xdyydy \Rightarrow 3x\,dx + 4y\,dx = 3x\,dy - y\,dy 3xdy4ydx=0 \Rightarrow 3x\,dy - 4y\,dx = 0

STEP 4

Rearrange the terms to clearly separate dxdx and dydy.
3xdy4ydx=0 3x\,dy - 4y\,dx = 0 4ydx+3xdy=0 \Rightarrow -4y\,dx + 3x\,dy = 0 M(x,y)dx+N(x,y)dy=0 \Rightarrow M(x, y)dx + N(x, y)dy = 0 where M(x,y)=4yM(x, y) = -4y and N(x,y)=3xN(x, y) = 3x.

STEP 5

Verify that M(x,y)M(x, y) and N(x,y)N(x, y) are homogeneous functions of the same degree.
Check if M(tx,ty)=tnM(x,y)M(tx, ty) = t^n M(x, y) and N(tx,ty)=tnN(x,y)N(tx, ty) = t^n N(x, y) for some degree nn.

STEP 6

Substitute txtx and tyty into M(x,y)M(x, y) and N(x,y)N(x, y) and check for homogeneity.
M(tx,ty)=4(ty)=t(4y)=t1M(x,y) M(tx, ty) = -4(ty) = t(-4y) = t^1 M(x, y) N(tx,ty)=3(tx)=t(3x)=t1N(x,y) N(tx, ty) = 3(tx) = t(3x) = t^1 N(x, y) Both MM and NN are homogeneous of degree 1.

STEP 7

Use the substitution y=vxy = vx to transform the ODE into a separable equation.
Let y=vxy = vx, then dy=vdx+xdvdy = vdx + xdv.

STEP 8

Substitute y=vxy = vx and dy=vdx+xdvdy = vdx + xdv into the ODE.
4(vx)dx+3x(vdx+xdv)=0 -4(vx)dx + 3x(vdx + xdv) = 0 4vdx+3vdx+3xdv=0 \Rightarrow -4vdx + 3vdx + 3xdv = 0 (4v+3v)dx+3xdv=0 \Rightarrow (-4v + 3v)dx + 3xdv = 0 vdx+3xdv=0 \Rightarrow -vdx + 3xdv = 0

STEP 9

Separate variables vv and xx.
dvdx=v3x \frac{dv}{dx} = \frac{v}{3x}

STEP 10

Integrate both sides with respect to xx.
dvv=13dxx \int \frac{dv}{v} = \frac{1}{3} \int \frac{dx}{x}

STEP 11

Perform the integration.
lnv=13lnx+C \ln|v| = \frac{1}{3}\ln|x| + C where CC is the constant of integration.

STEP 12

Exponentiate both sides to solve for vv.
elnv=e13lnx+C e^{\ln|v|} = e^{\frac{1}{3}\ln|x| + C} v=x13eC |v| = x^{\frac{1}{3}}e^C v=±x13eC v = \pm x^{\frac{1}{3}}e^C Let C1=±eCC_1 = \pm e^C to simplify.

STEP 13

Write vv in terms of xx and the constant C1C_1.
v=C1x13 v = C_1 x^{\frac{1}{3}}

STEP 14

Substitute back v=yxv = \frac{y}{x} to find the solution in terms of xx and yy.
yx=C1x13 \frac{y}{x} = C_1 x^{\frac{1}{3}}

STEP 15

Solve for yy.
y=C1x43 y = C_1 x^{\frac{4}{3}}

STEP 16

Write the final solution.
The general solution to the differential equation is:
y=C1x43 y = C_1 x^{\frac{4}{3}}
where C1C_1 is an arbitrary constant.

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