Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 90770 by ajfour last updated on 26/Apr/20

$$(ax+by+c)dx+(px+qy+r)dy=0$$

Answered by mr W last updated on 26/Apr/20

$$\frac{dy}{dx}=-\frac{ax+by+c}{px+qy+r}$$$$letx=u+u_0,y=v+v_0$$$$ax+by+c=a(u+u_0)+b(v+v_0)+c=au+bv+{au}_0+{bv}_0+c$$$$px+qy+r=p(u+u_0)+q(v+v_0)+r=pu+qv+{pu}_0+{qv}_0+r$$$$set{au}_0+{bv}_0+c=0$$$$set{pu}_0+{qv}_0+r=0$$$$\Rightarrow u_0=\frac{-cq+rb}{aq-bp}$$$$\Rightarrow v_0=\frac{-cp+ra}{aq-bp}$$$$\frac{dy}{dx}=\frac{dv}{du}=\frac{au+bv}{pu+qv}$$$$letv=ut$$$$\frac{dv}{du}=t+u\frac{dt}{du}$$$$t+u\frac{dt}{du}=\frac{a+bt}{p+qt}$$$$u\frac{dt}{du}=\frac{a+bt}{p+qt}-t=-\frac{{qt}^2+(p-b)t-a}{qt+p}$$$$\frac{qt+p}{{qt}^2+(p-b)t-a}dt=-\frac{du}{u}$$$$\int \frac{t+\frac{p}{q}}{t^2+(\frac{p}{q}-\frac{b}{q})t-\frac{a}{q}}dt=-\int \frac{du}{u}$$$$\int \frac{t+C}{t^2+At+B}dt=-\int \frac{du}{u}$$$$\frac{1}{2}\ln (t^2+At+B)+(C-\frac{A}{2})\int \frac{dt}{t^2+At+B}=-\int \frac{du}{u}$$$$\frac{1}{2}\ln (t^2+At+B)+\frac{2C-A}{\sqrt{4B-A^2}}{\tan }^{-1}\frac{2t+A}{\sqrt{4B-A^2}}=-\ln u+K$$$$(hereonlycase4B-A^2>0,forcases$$$$4B-A^2⩽0similarly)$$$$\Rightarrow \frac{1}{2}\ln [{\left(\frac{y-v_0}{x-u_0}\right)}^2+A\left(\frac{y-v_0}{x-u_0}\right)+B]+\frac{2C-A}{\sqrt{4B-A^2}}{\tan }^{-1}\frac{2\left(\frac{y-v_0}{x-u_0}\right)+A}{\sqrt{4B-A^2}}+\ln (x-u_0)=K$$

Commented by ajfour last updated on 26/Apr/20

$$ThanksSir,yourpresentation$$$$ofeverysolutionisbeyondall$$$$praise!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com