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Question Number 106774 by bemath last updated on 07/Aug/20

     ^(@bemath@)    (1)    3^x  + 3^((√x) ) = 90. find x ?     (2) x (dy/dx)−(1+x)y = xy^2

$$\:\:\:\:\overset{@\mathrm{bemath}@} {\:} \\ $$$$\:\left(\mathrm{1}\right)\:\:\:\:\mathrm{3}^{{x}} \:+\:\mathrm{3}^{\sqrt{{x}}\:} =\:\mathrm{90}.\:\mathrm{find}\:{x}\:?\: \\ $$$$\:\:\left(\mathrm{2}\right)\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}−\left(\mathrm{1}+\mathrm{x}\right)\mathrm{y}\:=\:\mathrm{xy}^{\mathrm{2}} \\ $$

Answered by john santu last updated on 07/Aug/20

(1) by observe x = 4 ⇒3^(4 )  + 3^(√4)  =  81 + 9 = 90. ^(@JS@)

$$\left(\mathrm{1}\right)\:\mathrm{by}\:\mathrm{observe}\:\mathrm{x}\:=\:\mathrm{4}\:\Rightarrow\mathrm{3}^{\mathrm{4}\:} \:+\:\mathrm{3}^{\sqrt{\mathrm{4}}} \:= \\ $$$$\mathrm{81}\:+\:\mathrm{9}\:=\:\mathrm{90}.\:\:^{@\mathrm{JS}@} \: \\ $$

Answered by bobhans last updated on 07/Aug/20

        ^(≻bobhans≺)   (dy/dx) − (((1+x)/x))y = y^2   (Bernoulli diff eq )  set u = y^(−1)  ⇒(du/dx) = −y^(−2)  (dy/dx) ; (dy/dx) =−y^2  (du/dx)  ⇔ −y^2  (du/dx)−(((1+x)/x))y=y^2   (du/dx) + (((1+x)/x)) u = −1 ; integrating factor   v(x)= e^(∫ (1+(1/x))dx) = e^(x+ln x)  = xe^x   u(x)= ((∫−xe^x  dx +C)/(xe^x )) = ((−xe^x +e^x +C)/(xe^x ))  (1/y) = ((−xe^x +e^x +C)/(xe^x )) ⇒ y = ((xe^x )/(C−xe^x +e^x ))

$$\:\:\:\:\:\:\:\:\:^{\succ\mathrm{bobhans}\prec} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:−\:\left(\frac{\mathrm{1}+\mathrm{x}}{\mathrm{x}}\right)\mathrm{y}\:=\:\mathrm{y}^{\mathrm{2}} \:\:\left(\mathrm{Bernoulli}\:\mathrm{diff}\:\mathrm{eq}\:\right) \\ $$$$\mathrm{set}\:\mathrm{u}\:=\:\mathrm{y}^{−\mathrm{1}} \:\Rightarrow\frac{\mathrm{du}}{\mathrm{dx}}\:=\:−\mathrm{y}^{−\mathrm{2}} \:\frac{\mathrm{dy}}{\mathrm{dx}}\:;\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=−\mathrm{y}^{\mathrm{2}} \:\frac{\mathrm{du}}{\mathrm{dx}} \\ $$$$\Leftrightarrow\:−\mathrm{y}^{\mathrm{2}} \:\frac{\mathrm{du}}{\mathrm{dx}}−\left(\frac{\mathrm{1}+\mathrm{x}}{\mathrm{x}}\right)\mathrm{y}=\mathrm{y}^{\mathrm{2}} \\ $$$$\frac{\mathrm{du}}{\mathrm{dx}}\:+\:\left(\frac{\mathrm{1}+\mathrm{x}}{\mathrm{x}}\right)\:\mathrm{u}\:=\:−\mathrm{1}\:;\:\mathrm{integrating}\:\mathrm{factor}\: \\ $$$$\mathrm{v}\left(\mathrm{x}\right)=\:\mathrm{e}^{\int\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{dx}} =\:\mathrm{e}^{\mathrm{x}+\mathrm{ln}\:\mathrm{x}} \:=\:\mathrm{xe}^{\mathrm{x}} \\ $$$$\mathrm{u}\left(\mathrm{x}\right)=\:\frac{\int−\mathrm{xe}^{\mathrm{x}} \:\mathrm{dx}\:+\mathrm{C}}{\mathrm{xe}^{\mathrm{x}} }\:=\:\frac{−\mathrm{xe}^{\mathrm{x}} +\mathrm{e}^{\mathrm{x}} +\mathrm{C}}{\mathrm{xe}^{\mathrm{x}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{y}}\:=\:\frac{−\mathrm{xe}^{\mathrm{x}} +\mathrm{e}^{\mathrm{x}} +\mathrm{C}}{\mathrm{xe}^{\mathrm{x}} }\:\Rightarrow\:\mathrm{y}\:=\:\frac{\mathrm{xe}^{\mathrm{x}} }{\mathrm{C}−\mathrm{xe}^{\mathrm{x}} +\mathrm{e}^{\mathrm{x}} }\: \\ $$

Answered by abdomathmax last updated on 07/Aug/20

3) xy^′ −(1+x)y =xy^2  ⇒(y^′ /y^2 )−(1+(1/x))(1/y)=1  let z =(1/y) ⇒z^′  =−(y^′ /y^2 )  e⇒−z^′ −(1+(1/x))z =1 ⇒z^′  +(1+(1/x))z =−1  he→z^′ =−(1+(1/x))z ⇒(z^′ /z) =−(1+(1/x)) ⇒  ln∣z∣ =−x−ln∣x∣ +c ⇒z = k e^(−x) ×(1/(∣x∣))  solution on{x /x>0} ⇒z =k (e^(−x) /x)  mvc mrthod→z^(′ )  =k^′  (e^(−x) /x) +k ×((−xe^(−x) −e^(−x) )/x^2 )  =k^′  (e^(−x) /x) −k  (e^(−x) /x) −k (e^(−x) /x^2 )  e ⇒ k^′  (e^(−x) /x) −k (e^(−x) /x)−k (e^(−x) /x^2 ) +(1+(1/x))k (e^(−x) /x) =−1  ⇒k^′   =−x e^x  ⇒k =−∫x e^x  dx +c  =−{ xe^x −∫ e^x  dx} +c=−(x−1)e^x  +c ⇒  z =(e^(−x) /x)( −(x−1)e^x +c) =−((x−1)/x) +((ce^(−x) )/x)  we hsvr y =(1/z) ⇒ y =(x/(1−x +ce^(−x) ))

$$\left.\mathrm{3}\right)\:\mathrm{xy}^{'} −\left(\mathrm{1}+\mathrm{x}\right)\mathrm{y}\:=\mathrm{xy}^{\mathrm{2}} \:\Rightarrow\frac{\mathrm{y}^{'} }{\mathrm{y}^{\mathrm{2}} }−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\frac{\mathrm{1}}{\mathrm{y}}=\mathrm{1} \\ $$$$\mathrm{let}\:\mathrm{z}\:=\frac{\mathrm{1}}{\mathrm{y}}\:\Rightarrow\mathrm{z}^{'} \:=−\frac{\mathrm{y}^{'} }{\mathrm{y}^{\mathrm{2}} } \\ $$$$\mathrm{e}\Rightarrow−\mathrm{z}^{'} −\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{z}\:=\mathrm{1}\:\Rightarrow\mathrm{z}^{'} \:+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{z}\:=−\mathrm{1} \\ $$$$\mathrm{he}\rightarrow\mathrm{z}^{'} =−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{z}\:\Rightarrow\frac{\mathrm{z}^{'} }{\mathrm{z}}\:=−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\:\Rightarrow \\ $$$$\mathrm{ln}\mid\mathrm{z}\mid\:=−\mathrm{x}−\mathrm{ln}\mid\mathrm{x}\mid\:+\mathrm{c}\:\Rightarrow\mathrm{z}\:=\:\mathrm{k}\:\mathrm{e}^{−\mathrm{x}} ×\frac{\mathrm{1}}{\mid\mathrm{x}\mid} \\ $$$$\mathrm{solution}\:\mathrm{on}\left\{\mathrm{x}\:/\mathrm{x}>\mathrm{0}\right\}\:\Rightarrow\mathrm{z}\:=\mathrm{k}\:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}} \\ $$$$\mathrm{mvc}\:\mathrm{mrthod}\rightarrow\mathrm{z}^{'\:} \:=\mathrm{k}^{'} \:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}}\:+\mathrm{k}\:×\frac{−\mathrm{xe}^{−\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\mathrm{k}^{'} \:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}}\:−\mathrm{k}\:\:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}}\:−\mathrm{k}\:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{e}\:\Rightarrow\:\mathrm{k}^{'} \:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}}\:−\mathrm{k}\:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}}−\mathrm{k}\:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} }\:+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{k}\:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}}\:=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{k}^{'} \:\:=−\mathrm{x}\:\mathrm{e}^{\mathrm{x}} \:\Rightarrow\mathrm{k}\:=−\int\mathrm{x}\:\mathrm{e}^{\mathrm{x}} \:\mathrm{dx}\:+\mathrm{c} \\ $$$$=−\left\{\:\mathrm{xe}^{\mathrm{x}} −\int\:\mathrm{e}^{\mathrm{x}} \:\mathrm{dx}\right\}\:+\mathrm{c}=−\left(\mathrm{x}−\mathrm{1}\right)\mathrm{e}^{\mathrm{x}} \:+\mathrm{c}\:\Rightarrow \\ $$$$\mathrm{z}\:=\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{x}}\left(\:−\left(\mathrm{x}−\mathrm{1}\right)\mathrm{e}^{\mathrm{x}} +\mathrm{c}\right)\:=−\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}\:+\frac{\mathrm{ce}^{−\mathrm{x}} }{\mathrm{x}} \\ $$$$\mathrm{we}\:\mathrm{hsvr}\:\mathrm{y}\:=\frac{\mathrm{1}}{\mathrm{z}}\:\Rightarrow\:\mathrm{y}\:=\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}\:+\mathrm{ce}^{−\mathrm{x}} } \\ $$

Answered by Dwaipayan Shikari last updated on 07/Aug/20

(dy/dx)−(((1+x))/x)y=y^2                 (1/y) (dy/dx)−((1+x)/x)=y                              {(1/y)=v ,   (dy/dx)v+y(dv/dx)=0  ,(1/y).(dy/dx)=−(1/v).(dv/dx)  −(1/v).(dv/dx)−((1+x)/x)=(1/v)  (dv/dx)+((1+x)/x)v=−1  I.F=e^(∫((1+x)/x)) =e^((logx+x)) =xe^x   v.xe^x =−∫xe^x   v.xe^x =−xe^x +e^x +C  v=−1+(1/x)+(C/x)e^(−x)   (1/y)=−1+(1/x)+(C/x)e^(−x)   (1/y)=((−xe^x +e^x +C)/(xe^x ))  y=((xe^x )/((e^x −xe^x +C)))

$$\frac{{dy}}{{dx}}−\frac{\left(\mathrm{1}+{x}\right)}{{x}}{y}={y}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\frac{\mathrm{1}}{{y}}\:\frac{{dy}}{{dx}}−\frac{\mathrm{1}+{x}}{{x}}={y}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\frac{\mathrm{1}}{{y}}={v}\:,\:\:\:\frac{{dy}}{{dx}}{v}+{y}\frac{{dv}}{{dx}}=\mathrm{0}\:\:,\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dx}}=−\frac{\mathrm{1}}{{v}}.\frac{{dv}}{{dx}}\right. \\ $$$$−\frac{\mathrm{1}}{{v}}.\frac{{dv}}{{dx}}−\frac{\mathrm{1}+{x}}{{x}}=\frac{\mathrm{1}}{{v}} \\ $$$$\frac{{dv}}{{dx}}+\frac{\mathrm{1}+{x}}{{x}}{v}=−\mathrm{1} \\ $$$${I}.{F}={e}^{\int\frac{\mathrm{1}+{x}}{{x}}} ={e}^{\left({logx}+{x}\right)} ={xe}^{{x}} \\ $$$${v}.{xe}^{{x}} =−\int{xe}^{{x}} \\ $$$${v}.{xe}^{{x}} =−{xe}^{{x}} +{e}^{{x}} +{C} \\ $$$${v}=−\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{{C}}{{x}}{e}^{−{x}} \\ $$$$\frac{\mathrm{1}}{{y}}=−\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{{C}}{{x}}{e}^{−{x}} \\ $$$$\frac{\mathrm{1}}{{y}}=\frac{−{xe}^{{x}} +{e}^{{x}} +{C}}{{xe}^{{x}} } \\ $$$${y}=\frac{{xe}^{{x}} }{\left({e}^{{x}} −{xe}^{{x}} +\mathrm{C}\right)} \\ $$

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