Question

Consider the following equation, $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
(i) If two particular solutions of given equation $u\left(x\right)$ and $v\left(x\right)$ are known, find the general solution of the same equation in terms of $u\left(x\right)$ and $v\left(x\right)$.
(ii) If $\alpha$ and $\beta$ are constants such that the linear combinations $\alpha \cdot u\left(x\right)+\beta \cdot v\left(x\right)$ is a solution of the given equation, find the relation between $\alpha$ and $\beta$.
(iii) If $w\left(x\right)$ is the third particular solution different from $u\left(x\right)$ and $v\left(x\right)$ then find the ratio $\frac{v\left(x\right)-u\left(x\right)}{w\left(x\right)-u\left(x\right)}$.

Answer 1

The following is a linear differential equation, whose solution is given by, $I.y=\int I.Q\left(x\right)dx+c$, where $I=e^{\int P\left(x\right)dx}$ is the integrating factor and '$c$' is the constant of integration.

Since $u\left(x\right)$ is a solution of the equation, we can write: $u^{\prime }\left(x\right)+P\left(x\right)u\left(x\right)=Q\left(x\right)$ ...(1) ....($u^{\prime }\left(x\right)$ is the first derivative of $u\left(x\right)$ wrt $x$).

Similarly, we can write: $v^{\prime }\left(x\right)+P\left(x\right)v\left(x\right)=Q\left(x\right)$ ...(2)

i) Using equation (1) and (2), we write $P\left(x\right)$ and $Q\left(x\right)$ in terms of $u\left(x\right)$ and $v\left(x\right)$. Subtracting (2) from (1), we get $P\left(x\right)=\frac{v^{\prime }\left(x\right)-u^{\prime }\left(x\right)}{u\left(x\right)-v\left(x\right)}$.

Similarly, $Q\left(x\right)=\frac{v^{\prime }\left(x\right)u\left(x\right)-u^{\prime }\left(x\right)v\left(x\right)}{u\left(x\right)-v\left(x\right)}$

Therefore, $I=e^{1234\int \frac{-\left(u^{\prime }\right(x)-v^{\prime }(x\left)\right)}{u\left(x\right)-v\left(x\right)}dx}=e^{ln\left(u\right(x)-v(x){)}^{-1}}=\frac{1}{u\left(x\right)-v\left(x\right)}$

Hence, the general solution is, $(u\left(x\right)-v\left(x\right)).\int \frac{-(u^{\prime }\left(x\right)v\left(x\right)-v^{\prime }\left(x\right)u\left(x\right))}{\left(u\right(x)-v(x){)}^2}dx+c.\left(u\right(x)-v(x\left)\right)$

ii.) Given $\alpha u\left(x\right)+\beta v\left(x\right)$ is also a solution.

Hence, we write,

$\left(\alpha u^{\prime }\right(x)+\beta v^{\prime }(x\left)\right)+P\left(x\right).\left(\alpha u\right(x)+\beta v(x\left)\right)=Q\left(x\right)$.

Rearranging the terms, we get: $\alpha \left(u^{\prime }\right(x)+P(x\left)u\right(x\left)\right)+\beta \left(v^{\prime }\right(x)+P(x\left)v\right(x\left)\right)=Q\left(x\right)$

Using equation (1) and (2), we get: $\alpha Q\left(x\right)+\beta Q\left(x\right)=Q\left(x\right)$

We can say: $\alpha +\beta =1$

iii.) Let $c_1,c_2$ and $c_3$ be the constants of integration for $u\left(x\right),v\left(x\right)$ and $w\left(x\right)$ respectively.

$v\left(x\right)-u\left(x\right)=$ $\frac{c_2-c_1}{I}$, where $I$ is the integration factor.

Similarly, $w\left(x\right)-u\left(x\right)=$ $\frac{c_3-c_1}{I}$.

Dividing, both the equations, we get $\frac{c_2-c_1}{c_3-c_1}=k\left(\text{say}\right)$.

Hence, the ratio is a constant.