solving a differential equation involving hyperbolic functions

Question

Given:

${u}^{'}(x)- cosh(x)u(x) = \sqrt{1+sinh^{2}(x)}$

and

$u(0)=e-1$

What I have so far:

I have set $\sqrt{1+sinh^{2}(x)} = 0$, in order to solve for the homogeous part of the solution

$\Rightarrow {u}^{'}(x)- cosh(x)u(x) = 0$

$\Rightarrow du(x)- cosh(x)u(x)dx = 0$

$\Rightarrow du(x) = cosh(x)u(x)dx$

$\Rightarrow \frac{du(x)}{u(x)} = cosh(x)dx$

Then integrated both sides:

$\Rightarrow \int\frac{du(x)}{u(x)} = \int cosh(x)dx$

$\Rightarrow ln|u(x)| = sinh(x) + C$

$\Rightarrow e^{ln|u(x)|} = e^{sinh(x)+C}$

$\Rightarrow u_{h}(x) = C_{1}e^{sinh(x)}$

Now the particular solution:

$u_{p}(x) = C_{2}$

substituting the constant back into the equation:

$\Rightarrow C_{2}=\frac{\sqrt{1+sinh^{2}(x)}}{cosh(x)}$

the solution is then:

$\Rightarrow u(x) = C_{1}e^{sinh(x)}-\frac{\sqrt{1+sinh^{2}(x)}}{cosh(x)}$

Now substituting the initial value $u(0) = e-1$

I got $C_{1} = e$

Thus the equation is:

$u(x) = e^{sinh(x)+1}-\frac{\sqrt{1+sinh^{2}(x)}}{cosh(x)}$

My questions:

1. Am I allowed to set the term $\sqrt{1+sinh^{2}(x)}$ to 0, even though it is not a constant?

2. Is it okay to set the particular solution to a constant like i did in my solution?

3. Is the answer correct?

Answer 1

Yes for the first question since you solved the homogeneous DE $${u}^{'}(x)- \cosh(x)u(x) = 0$$ Your answer is correct. Note that $1+\sinh ^2 x=\cosh^2 x$.

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Note that you can simplify the DE just susbtitute: $$\dfrac {du}{dx}=\dfrac {du}{d \sinh x }\dfrac {d \sinh x}{dx}=\cosh x \dfrac {du}{d \sinh x}$$ Then $${u}^{'}(x)- \cosh(x)u(x) = \sqrt{1+\sinh^{2}(x)}$$ $$u'(\sinh x)- u(\sinh x) =1$$ This DE is separable.

Answer 2

Observing that $\sqrt{1+\sinh^2x}=\cosh x$ you can write the DE as $$ u'-\cosh(x)(u+1)=0 $$ which is directly separable, giving $$ u(x)+1=e^{\sinh x}(u(0)+1) $$