I am totally new to Differential Equations (first week), and am stuck regarding the following question:

$$y'=y(1+xy^4),\quad y(0) = 1$$


a. $\frac{1}{y}=\sqrt[4]{\frac{1}{4}-x+\frac{3}{4}e^{-4x}}$

b. $y=\frac{1}{\sqrt[4]{1-x}}$

c. $\frac{1}{y}=\sqrt[4]{\frac{5}{4}e^{-4x}-\frac{1}{4}-x}$

d. $y=\frac{1}{4}-x+\frac{3}{4}e^{-4x}$

e. $\frac{1}{y}=\sqrt[4]{\frac{1}{4}+x+\frac{3}{4}e^{-4x}}$

f. There is no solution

I have tried separating $x$ and $y$ on either side of the equals mark, but failed. I have tried simply integrating both sides, but the outcome does not allow me to isolate $y$. Thus I am stuck, and I am not sure what I can do to resolve this. Could anybody please help out? Many thanks in advance!

  • 1
    $\begingroup$ Just replace each solution in the equation and see if one of them match. $\endgroup$ – Surb Jul 10 at 20:23
  • $\begingroup$ That's not a bad idea, but I would still like to understand the methodology of actually resolving this. $\endgroup$ – dalta Jul 10 at 20:34
  • $\begingroup$ I'm not sure that it's possible to solve it algebraically. It's not linear and I don't see any obvious substitution... so probably it's not. $\endgroup$ – Surb Jul 10 at 20:42

Based on the fact that this is your first week I'm very sure that the point of this problem is for you to take the derivative of the choices and substitute them in the original question. I guess you are asking because your are interested. Interest should not be discouraged.

This question is solvable using the methods 'u-sub' and `Integrating Factors'. Its simple. Clean up the expression to get:


Let $$u = \frac{1}{y^4}\ (*)$$ so $u' = \frac{du}{dx} = -4\frac{y'}{y^5}$.

Now the original question becomes:

$$-\frac{u'}{4}-u = x$$

$$u'+4u = -4x$$

This is where you use `integrating factors'. Let $\mu = e^{\int 4 dx} = e^{4x}$ and multiply both sides with it.

$$e^{4x}u'+4e^{4x}u = -4xe^{4x}$$


$$ (ue^{4x})' = -4xe^{4x}$$

Integrate both sides to get $u$

$$ u e^{4x} = -4\int xe^{4x} dx$$

Here you use integration by parts. Get $u$ from here and then use equation (*) to get your answer, there will be an undetermined integration factor you determine by using the boundary condition. But that is easy.

  • $\begingroup$ Amazing, thank you! $\endgroup$ – dalta Jul 10 at 21:18

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