Menu Close

Solve-this-differential-equation-a-L-a-b-L-b-L-where-L-0-a-a-2-sin-2-t-b-2-cos-2-t-dt-

Tinku Tara 0 Comments
Pin




Question Number 161820 by HongKing last updated on 22/Dec/21
Solve this differential equation: a ((∂L(α))/∂a) + b ((∂L(α))/∂b) = L(α) where: L(α) = ∫_( 0) ^( a) (√(a^2 sin^2 (t) + b^2 cos^2 (t))) dt
$$\mathrm{Solve}\:\mathrm{this}\:\mathrm{differential}\:\mathrm{equation}: \\ $$$${a}\:\frac{\partial{L}\left(\alpha\right)}{\partial{a}}\:+\:{b}\:\frac{\partial{L}\left(\alpha\right)}{\partial{b}}\:=\:{L}\left(\alpha\right) \\ $$$$\mathrm{where}:\:{L}\left(\alpha\right)\:=\:\underset{\:\mathrm{0}} {\overset{\:\boldsymbol{{a}}} {\int}}\sqrt{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({t}\right)\:+\:{b}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({t}\right)}\:{dt} \\ $$
Answered by aleks041103 last updated on 23/Dec/21
L(x)=∫_0 ^x (√(a^2 sin^2 t+b^2 cos^2 t))dt a(∂L/∂a)=∫_0 ^x (∂/∂a)(√(a^2 sin^2 t+b^2 cos^2 t)) dt= =∫_0 ^x ((a^2 sin^2 t)/( (√(a^2 sin^2 t+b^2 cos^2 t))))dt by analogy b(∂L/∂b)=∫_0 ^x ((b^2 cos^2 t)/( (√(a^2 sin^2 t+b^2 cos^2 t))))dt then a(∂L/∂a)+b(∂L/∂b)=∫_0 ^x ((a^2 sin^2 t+b^2 cos^2 t)/( (√(a^2 sin^2 t+b^2 cos^2 t))))dt= =∫_0 ^x (√(a^2 sin^2 t+b^2 cos^2 t))dt=L ⇒this is the general solution
$${L}\left({x}\right)=\underset{\mathrm{0}} {\overset{{x}} {\int}}\sqrt{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}+{b}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}}{dt} \\ $$$${a}\frac{\partial{L}}{\partial{a}}=\underset{\mathrm{0}} {\overset{{x}} {\int}}\frac{\partial}{\partial{a}}\sqrt{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}+{b}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}}\:{dt}= \\ $$$$=\int_{\mathrm{0}} ^{{x}} \frac{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}}{\:\sqrt{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}+{b}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}}}{dt} \\ $$$${by}\:{analogy} \\ $$$${b}\frac{\partial{L}}{\partial{b}}=\int_{\mathrm{0}} ^{{x}} \frac{{b}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}}{\:\sqrt{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}+{b}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}}}{dt} \\ $$$${then} \\ $$$${a}\frac{\partial{L}}{\partial{a}}+{b}\frac{\partial{L}}{\partial{b}}=\int_{\mathrm{0}} ^{{x}} \frac{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}+{b}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}}{\:\sqrt{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}+{b}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}}}{dt}= \\ $$$$=\int_{\mathrm{0}} ^{{x}} \sqrt{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}+{b}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}}{dt}={L} \\ $$$$\Rightarrow{this}\:{is}\:{the}\:{general}\:{solution} \\ $$
Commented by HongKing last updated on 25/Dec/21
thank you so much my dear Sir cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *