Question

Write the equation in the form y′=f(y/x) then use the substitution y=xu to find an implicit general solution. Then solve the initial value problem.


y′=5y2+4x2xy, y(1)=3


The resulting differential equation in x x and u u can be written as xu′= which is separable.

Separating variables we arrive at du=dx/x

Separating variables and and simplifying the solution can be written in the form u2+1=Cf(x) where C is an arbitrary constant and f(x)= .

Transforming back into the variables x and y and using the initial condition to find C we find the explicit solution of the initial value problem is y = .

Answer 1

Looks like the ODE is supposed to be

$y'=\dfrac{5y^2+4x^2}{xy}$

In that case, let $y(x)=xu(x)$, so that $y'=xu'+u$. Substitute these into the ODE to get

$xu'+u=\dfrac{5x^2u^2+4x^2}{x^2u}$

$xu'=\dfrac{4u^2+4}u$

which is separable as

$\dfrac u{u^2+1}\,\mathrm du=4x\,\mathrm dx$

Integrate both sides to get

$\dfrac12\ln(u^2+1)=2x^2+C$

We can solve for $u^2$:

$\ln(u^2+1)=4x^2+C$

$u^2+1=Ce^{4x^2}$

$u^2=Ce^{4x^2}-1$

Replace $u=\frac yx$ to get the general solution,

$\dfrac{y^2}{x^2}=Ce^{4x^2}-1$

Given that $y(1)=3$, we find

$9=Ce^4-1\implies C=10e^{-4}$

so the particular solution is

$\dfrac{y^2}{x^2}=10e^{4(x^2-1)}-1$

$\implies y=\pm x\sqrt{10e^{4(x^2-1)}-1}$

Answer 2

The solution of this differential equation is $y = \pm x \cdot \sqrt{49\cdot x^{10}-0.8}$.

How to solve a non-linear differential equation

Now we proceed to write the differential equation in the form $y'=f\left(\frac{y}{x} \right)$:

$y' = \frac{5\cdot y^{2}+4\cdot x^{2}}{x\cdot y}$ (1)

$y' = 5\cdot \frac{y}{x}+4\cdot \frac{x}{y}$

If we know that $u = \frac{y}{x}$, then we have the following expression and separate the variables:

$x\cdot u' = 5\cdot u + \frac{4}{u}$

$\int {\frac{u}{5\cdot u^{2}+4} } \, du = \int {\frac{dx}{x} }$

If we apply the substitution $k = 5\cdot u^{2}+4$, then we have the following integral:

$\frac{1}{10}\int {\frac{dk}{k} } = \int {\frac{dx}{x} }$ (2)

And we solve it:

$\frac{1}{10}\cdot \ln k = \ln x + C$

$k = C\cdot e^{x^{10}}$

$5\cdot u^{2}+4 = C\cdot x^{10}$

$5\cdot \frac{y^{2}}{x^{2}} + 4 = C\cdot x^{10}$

$5\cdot y^{2} = C\cdot x^{12}-4\cdot x^{2}$

Where $C$ is the integration constant.

If we know that $x = 1$ and $y = 3$, then the integration constant is:

$5\cdot (3)^{2} = C\cdot (1)^{12}-4\cdot (1)^{2}$

$C = 49$

And the solution of this differential equation is:

$y = \pm x \cdot \sqrt{49\cdot x^{10}-0.8}$ (3)

The solution of this differential equation is $y = \pm x \cdot \sqrt{49\cdot x^{10}-0.8}$. $\blacksquare$

Remark

The statement present mistakes and is poorly formatted. Correct form is described below:

Write the equation in the form $y' = f\left(\frac{y}{x} \right)$, then use the substitution $y= x\cdot u$ to find an implicit general solution. Then solve the initial value problem:

$y' = \frac{5\cdot y^{2}+4\cdot x^{2}}{x\cdot y}$ (1)

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