aspifsGak5u
2022-01-21
Answered

Find the general solution of these first order differential equations

$\left(1\u2013x\right){y}^{\prime}={y}^{2}$

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usumbiix

Answered 2022-01-21
Author has **33** answers

Given that:

$\left(1\u2013x\right){y}^{\prime}={y}^{2}$

Consider

$(1-x){y}^{\prime}={y}^{2}$

$(1-x)\frac{dy}{dx}={y}^{2}$

$\frac{dy}{{y}^{2}}=\frac{1}{1-x}dx$

Integrate on both sides

$\int \frac{1}{{y}^{2}}dy=\int \frac{1}{1-x}dx$

Let$u=1-x\text{}\Rightarrow \text{}\frac{du}{dx}=-1$

$-du=dx$

Hence we have

$\int \frac{1}{{y}^{2}}dy=-\int \frac{1}{u}du$

$[\because \int {y}^{n}dn=\frac{{y}^{n+1}}{n+1}+c]$

$\frac{{y}^{-2+1}}{-2+1}=-\mathrm{ln}(1-x)+c$

$-\frac{1}{y}=-\mathrm{ln}(1-x)+c$

$\Rightarrow y=\frac{1}{\mathrm{ln}(1-x)-c}$

$\therefore$ The general solution is

$y=\frac{1}{\mathrm{ln}(1-x)-c}$

Consider

Integrate on both sides

Let

Hence we have

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asked 2022-06-11

I would like to solve:

$y-{y}^{\prime}x-{y}^{\prime 2}=0$

In order to do so, we let ${y}^{\prime}=t$, and we assume x as a function of t. Now, we take derivative with respect to t from the differential equation, and obtain

$\frac{dy}{dt}-x-t\frac{dx}{dt}-2t=0$

By the chain rule, we have: $dy/dt=tdx/dt$. So, the above simplifies to

$x=-2t$

That is, we have: $x=-2dy/dx$. Thus, we obtain

$y=-\frac{{x}^{2}}{4}+C$

Now, if we want to verify the solution, it turns out that C must be zero, in other words, $y=-{x}^{2}/4$ satisfies the original differential equation.

I have two questions:

1) What happens to the integration constant? That is, what is the general solution of the differential equation?

2) If we try to solve this differential equation with Mathematica, we obtain

$y={C}_{1}x+{C}_{1}^{2}$,

which has a different form from the analytical approach. How can we also produce this result analytically?

$y-{y}^{\prime}x-{y}^{\prime 2}=0$

In order to do so, we let ${y}^{\prime}=t$, and we assume x as a function of t. Now, we take derivative with respect to t from the differential equation, and obtain

$\frac{dy}{dt}-x-t\frac{dx}{dt}-2t=0$

By the chain rule, we have: $dy/dt=tdx/dt$. So, the above simplifies to

$x=-2t$

That is, we have: $x=-2dy/dx$. Thus, we obtain

$y=-\frac{{x}^{2}}{4}+C$

Now, if we want to verify the solution, it turns out that C must be zero, in other words, $y=-{x}^{2}/4$ satisfies the original differential equation.

I have two questions:

1) What happens to the integration constant? That is, what is the general solution of the differential equation?

2) If we try to solve this differential equation with Mathematica, we obtain

$y={C}_{1}x+{C}_{1}^{2}$,

which has a different form from the analytical approach. How can we also produce this result analytically?

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