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You an immediately translate this integral equation into an easy second order linear ODE on $[0,1]$ with boundary conditions $x(0)=0$, $x(1)=1$. (I'm not going to do it for you). Just note that your integrand $ts-\min(t,s)$ is the Green function of the Laplacian in dimension $1$ (i.e. the second derivative) with Dirichlet boundary condition: one has $$\int_0^1\big(sr-\min(t,s)\big)x(t)ds=u(t)$$ if and only $$\cases{\ddot u(t)=x(t)\\u(0)=0\\u(1)=0\,.}$$ (The easy computation to check the latter starts writing $\int_0^1=\int_0^t+\int_t^1$; then pull the $t$-factors out of the integrals and derive (...). After that, you may want to precise further the above claim in the various functional settings; however for your eigenvalue problem everything is necessarily classic and smooth).

You can immediately translate this integral equation into an easy second order linear ODE on $[0,1]$ with boundary conditions $x(0)=0$, $x(1)=1$. (I'm not going to do it for you). Just note that your integrand $ts-\min(t,s)$ is the Green function of the Laplacian in dimension $1$ (i.e. the second derivative) with Dirichlet boundary condition: one has $$\int_0^1\big(sr-\min(t,s)\big)x(t)ds=u(t)$$ if and only $$\cases{\ddot u(t)=x(t)\\u(0)=0\\u(1)=0\,.}$$ (The easy computation to check the latter fact starts writing $\int_0^1=\int_0^t+\int_t^1$; then pull the $t$-factors out of the integrals and derive (...). After that, you may want to precise further the above claim in the various functional settings; however for your eigenvalue problem everything is necessarily classic and smooth).

You an immediately translate this integral equation into an easy second order linear ODE on $[0,1]$ with boundary conditions $x(0)=0$, $x(1)=1$. (I'm not going to do it for you). Just note that your integrand $ts-\min(t,s)$ is the Green function of the Laplacian in dimension $1$ (i.e. the second derivative) with Dirichlet boundary condition: one has $$\int_0^1\big(sr-\min(t,s)\big)x(t)ds=u(t)$$ if and only $$\cases{\ddot u(t)=x(t)\\u(0)=0\\u(1)=0\,.}$$ (The easy computation to check the latter starts writing $\int_0^1=\int_0^t+\int_t^1$; then pull the $t$-factors out of the integrals and derive (...). After that, you may want to precise further the above claim in the various functional setting; however for your eigenvalue problem everything is necessarily classic and smooth).

You an immediately translate this integral equation into an easy second order linear ODE on $[0,1]$ with boundary conditions $x(0)=0$, $x(1)=1$. (I'm not going to do it for you). Just note that your integrand $ts-\min(t,s)$ is the Green function of the Laplacian in dimension $1$ (i.e. the second derivative) with Dirichlet boundary condition: one has $$\int_0^1\big(sr-\min(t,s)\big)x(t)ds=u(t)$$ if and only $$\cases{\ddot u(t)=x(t)\\u(0)=0\\u(1)=0\,.}$$ (The easy computation to check the latter starts writing $\int_0^1=\int_0^t+\int_t^1$; then pull the $t$-factors out of the integrals and derive (...). After that, you may want to precise further the above claim in the various functional settings; however for your eigenvalue problem everything is necessarily classic and smooth).

You an immediately translate this integral equation into an easy second order linear ODE on $[0,1]$ with null boundary conditions (I'm not going to do it for you). Just note that your integrand $ts-\min(t,s)$ is the Green function of the Laplacian in dimension $1$ (i.e. the second derivative) with Dirichlet boundary condition: one has $$\int_0^1\big(sr-\min(t,s)\big)x(t)ds=u(t)$$ if and only $$\cases{\ddot u(t)=x(t)\\u(0)=0\\u(1)=0\,.}$$ (The easy computation to check this starts writing $\int_0^1=\int_0^t+\int_t^1$; then pull the $t$-factors out of the integrals and derive (...). After that, you may want to precise further the above claim in the various functional setting; however for your eigenvalue problem everything is necessarily classic and smooth).

You an immediately translate this integral equation into an easy second order linear ODE on $[0,1]$ with boundary conditions $x(0)=0$, $x(1)=1$. (I'm not going to do it for you). Just note that your integrand $ts-\min(t,s)$ is the Green function of the Laplacian in dimension $1$ (i.e. the second derivative) with Dirichlet boundary condition: one has $$\int_0^1\big(sr-\min(t,s)\big)x(t)ds=u(t)$$ if and only $$\cases{\ddot u(t)=x(t)\\u(0)=0\\u(1)=0\,.}$$ (The easy computation to check the latter starts writing $\int_0^1=\int_0^t+\int_t^1$; then pull the $t$-factors out of the integrals and derive (...). After that, you may want to precise further the above claim in the various functional setting; however for your eigenvalue problem everything is necessarily classic and smooth).