$(x+y)\frac{dy}{dx} -(4x+y)=0, y(1)=2$

Question

Hi I am stuck solving this problem: 'Employ a change of variables $z=\frac{y}{x}$ to solve the differential equation: $$(x+y)\frac{dy}{dx} -(4x+y)=0, y(1)=2$$

Enter the result $y(3)$.

I have done the substitution and then integrated by I cannot get an explicit expression for y, which leaves me confused as to how I can find $y(3)$. Please help!

Answer 1

Starting from

$$(x+y)\frac{dy}{dx} -(4x+y)=0$$

rewrite as

$$\frac{dy}{dx}=\frac{4x+4}{x+y}=\frac{4+\frac xy}{1+\frac xy}$$

then with the substitution $z=\frac{y}{x}$ we have $\frac{dy}{dx}=z+x\frac{dz}{dx}$ so

$$z+x\frac{dz}{dx}=\frac{4+z}{1+z} \implies x\frac{dz}{dx}=\frac{4-z^2}{1+z}$$

which we can rewrite as

$$\frac{1+z}{4-z^2}dz=\frac{1}{x}dx$$

where a partial fraction decomposition of the LHS forms

$$\frac{1+z}{4-z^2}=\frac{-1}{4(z+2)}-\frac{3}{4(z-2)}$$

hence

$$-\frac{1}{4}\left(\frac{1}{z+2}+\frac{3}{z-2}\right)dz=\frac{1}{x}dx$$

whereby integrating both sides forms

$$-\frac{1}{4}\Big(\ln|z+2|+3\ln|z-2|\Big)=\ln|x|+C$$ or $$\ln|z+2|+3\ln|z-2|=-4\ln|x|+C$$ which can be rewritten as $$(z+2)(z-2)^3=Cx^{-4} \implies C=x^4(z+2)(z-2)^3$$ therefore substituting $z=\frac{y}{x}$ $$C=x^4\left(\frac{y}{x}+2\right)\left(\frac{y}{x}-2\right)^3$$ the initial/boundary condition of $y(1)=2$ forms $$C=1(4)(0)=0$$ hence since $x\neq 0$ we see that $$(y+2x)(y-2x)^3=0$$ which implies $$y=-2x,\quad y=2x$$ where $y=-2x$ cannot be the solution since $y(1)=2$. Therefore, $y=2x$ is the solution and $y(3)=6$.

Answer 2

Taking $z = \frac{y}{x}$ yields $y=xz$, and hence $\frac{dy}{dx} = x\frac{dz}{dx}+z$. From this, we find by substitution \begin{align*} (x+xz)(x\frac{dz}{dx}+z) - (4x+xz) &= 0 \\ \text{ divide by } x, \text{ and simplify}\\ x \frac{dz}{dx} +z +zx \frac{dz}{dx} +z^2 -4-z &= 0 \\ \frac{dz}{dx}x(1+z) &= 4-z^2 \\ \frac{dz(1+z)}{4-z^2} &= \frac{dx}{x} \\ dz \left( \frac{3}{4(2-z)} - \frac{1}{4(2+z)} \right) &= \frac{dx}{x} \end{align*} Solving this differential equation, we get for some constant $K$, $$\frac{1}{|(z+2)(z-2)^3|} = Kx^4.$$

Substituting $z = \frac{y}{x}$ again, we get after simplifying $$ 1 = K|y^2-4x^2|(y-2x)^2 $$

Using the information that $y(1)=2$, we should be able to find $K$, but we get the following. \begin{align*} 1 &= K|2^2-4 \cdot 1^2|(2-2 \cdot 1)^2 \\ 1 &= 0 \end{align*} Are you sure you copied the condition $y(1)=2$ correctly? There is definetily some mistake in your formulation of the problem anyway.

The next step would be to solve the polynomial equation $$1 = K|y^2-4 \cdot 3^2|(y-2 \cdot 3)^2$$ Finding $y$ that fits this will then be the solution to your problem.