# Solving the Bernoulli D.E

#### bergausstein

##### Active member
Solve the ff:

$\displaystyle\frac{dy}{dx}-y=xy^5$

$\displaystyle\frac{dy}{dx}-\frac{y}{x}=-\frac{5}{2}x^2y^3$

can you help start solving these problems? thanks!

#### MarkFL

Staff member
These are Bernoulli equations. Such an equation may be written in the form:

$$\displaystyle \frac{dy}{dx}+P(x)y=Q(x)y^n$$

Use of the substitution:

$$\displaystyle v=y^{1-n}$$

will transform the ODE into a linear equation.

#### bergausstein

##### Active member

the n=5

$\displaystyle v=y^{1-5}=\frac{1}{y^4}$

$\displaystyle y=(\frac{1}{v})^{\frac{1}{4}}$

$\displaystyle y'=-\frac{1}{4}v^{-\frac{5}{4}}\frac{dv}{dx}$

substitute,

$\displaystyle y=(\frac{1}{v})^{\frac{1}{4}}$ and $\displaystyle y'=-\frac{1}{4}v^{-\frac{5}{4}}\frac{dv}{dx}$ to the original D.E

$\displaystyle -\frac{1}{4}v^{-\frac{5}{4}}\frac{dv}{dx}-\frac{1}{v^{\frac{1}{4}}}=x\frac{1}{v^{\frac{5}{4}}}$

multiplying both sides by $v^{\frac{5}{$}}$and -4.$\displaystyle \frac{dv}{dx}+4v=-4x$the integrating factor is$e^{4x}\displaystyle e^{4x}\frac{dv}{dx}+4ve^{4x}=-4xe^{4x}\displaystyle \int D_x(e^{4x}v)=\int -4xe^{4x}$the rhs using by parts$\displaystyle e^{4x}v= -e^{4x}+\frac{1}{4}e^{4x}+c$dividing both sides by$ e^{4x}\displaystyle v= -1+\frac{1}{4}+\frac{c}{e^{4x}}$now since$v=\frac{1}{y^4}\displaystyle\frac{1}{y^4}=\frac{c}{e^{4x}}-\frac{3}{4}$--->> this would be my implicit general solution is my answer correct? Last edited: #### MarkFL ##### Administrator Staff member We are given: $$\displaystyle \frac{dy}{dx}-y=xy^5$$ Use the substitution: $$\displaystyle v=y^{-4}\implies\frac{dv}{dx}= -4y^{-5}\frac{dy}{dx}$$ Divide the original ODE by$y^5$observing we are eliminating the trivial solution$y\equiv0$. We obtain: $$\displaystyle y^{-5}\frac{dy}{dx}-y^{-4}=x$$ Applying the substitution we obtain: $$\displaystyle \frac{dv}{dx}+4v=-4x$$ Multiply through by$e^{4x}: $$\displaystyle e^{4x}\frac{dv}{dx}+4e^{4x}v=-4xe^{4x}$$ This becomes: $$\displaystyle \frac{d}{dx}\left(e^{4x}v \right)=-4xe^{4x}$$ Integrating (using IBP on the right (which you did incorrectly)), we obtain: $$\displaystyle e^{4x}v=\frac{1}{4}e^{4x}\left(1-4x \right)+C$$ $$\displaystyle v=\frac{1}{4}\left(1-4x \right)+Ce^{-4x}$$ $$\displaystyle y^{-4}=\frac{1}{4}\left(1-4x \right)+Ce^{-4x}$$ Thus the solution may be given implicitly by: $$\displaystyle y^4=\frac{4}{1-4x+Ce^{-4x}}$$ Note: the trivial solution we eliminated is not obtainable by the implicit solution gave above. #### bergausstein ##### Active member can you tell me where's my mistake in my solution above? #### Prove It ##### Well-known member MHB Math Helper can you tell me where's my mistake in my solution above? Edit: There is a small mistake in the integration. Sorry. \displaystyle \begin{align*} \int{-4x\,e^{4x}\,dx} = -4x\,e^{4x} + \frac{1}{4}e^{4x} + C \end{align*} Last edited: #### bergausstein ##### Active member\displaystyle e^{4x}v= -xe^{4x}+\frac{1}{4}e^{4x}+c\displaystyle v= -x+\frac{1}{4}+\frac{c}{e^{4x}}$my final answer should be$\displaystyle\frac{1}{y^4}=\frac{c}{e^{4x}}-x+\frac{1}{4}$am I right? #### Prove It ##### Well-known member MHB Math Helper$\displaystyle e^{4x}v= -xe^{4x}+\frac{1}{4}e^{4x}+c\displaystyle v= -x+\frac{1}{4}+\frac{c}{e^{4x}}$my final answer should be$\displaystyle\frac{1}{y^4}=\frac{c}{e^{4x}}-x+\frac{1}{4}$am I right? Please re-read my last post. You keep missing the factor of -4 on the first term. #### bergausstein ##### Active member Please re-read my last post. You keep missing the factor of -4 on the first term.$\displaystyle\int -4xe^{4x}dx=-4\int xe^{4x}$let$u=x$and$du=dx$let$dv=e^{4x}dx$now$v=\frac{1}{4}e^{4x}uv-\int vdu\frac{1}{4}xe^{4x}-\frac{1}{4}\int e^{4x}dx-4\left(\frac{1}{4}xe^{4x}-\frac{1}{4}\frac{1}{4}e^{4x}+c\right)= -xe^{4x}+\frac{1}{4}e^{4x}+c$isn't correct? #### Prove It ##### Well-known member MHB Math Helper$\displaystyle\int -4xe^{4x}dx=-4\int xe^{4x}$let$u=x$and$du=dx$let$dv=e^{4x}dx$now$v=\frac{1}{4}e^{4x}uv-\int vdu\frac{1}{4}xe^{4x}-\frac{1}{4}\int e^{4x}dx-4\left(\frac{1}{4}xe^{4x}-\frac{1}{4}\frac{1}{4}e^{4x}+c\right)= -xe^{4x}+\frac{1}{4}e^{4x}+c\$

isn't correct?
Never mind, what you have is correct. I forgot the 1/4 factor in the v term.

I think this many mistakes means it's time for bed >_<